The Visual C++ compiler in the Microsoft Visual Studio 2010 September Community Technology Preview (CTP) contains support for four C++0x language features, namely lambdas, auto, static_assert, and rvalue references. Today, I'm going to explain the first three features in detail. (And I'm going to dedicate an entire post to explaining rvalue references in the near future, simply because explaining them now would double the length of this already very long post.)

First, a few quick things:

1. Today's post is brought to you by Stephan T. Lavavej, Visual C++ Libraries Developer, and the letters C, A, and T. Note that as a libraries dev, I didn't implement these features. That was the work of Jonathan Caves, front-end compiler dev, voting Standardization Committee member, and all-around ninja.

2. I refer to the Visual C++ compiler in VS 2010 as VC10 (VS 2008 contained VC9, VS 2005 contained VC8, etc. - 10 is not short for 2010).

3. C++0x refers to the upcoming C++ Standard, which is still being drafted. (The Standardization Committee hopes that they'll be finished in 2009, making it C++09; the joke is that if it slips to 2010 or later, the 'x' will be hexadecimal.) C++98 and C++03 refer to the current C++ Standard. (Without going into a history lecture here, the 2003 C++ Standard was merely a "service pack" for the original 1998 C++ Standard, and most people can disregard the differences. C++03 and C++0x are totally different, despite appearances.)

4. I'd like to thank the Standardization Committee for developing these wonderfully useful and well-crafted features. They also make important documents available on their website:

Working Draft, Standard for Programming Language C++: http://www.open-std.org/JTC1/SC22/WG21/docs/papers/2011/n3242.pdf

5. There are always bugs (although hopefully not too many), which is the whole point of the CTP. Please report bugs to us via Microsoft Connect.

Now, let's look at the features!

lambdas

In C++0x, "lambda expressions" implicitly define and construct unnamed function objects, which then behave like handwritten function objects. This is the "Hello, World" lambda:

C:\Temp>type meow.cpp

#include <algorithm>

#include <iostream>

#include <ostream>

#include <vector>

using namespace std;

int main() {

vector<int> v;

for (int i = 0; i < 10; ++i) {

v.push_back(i);

}

for_each(v.begin(), v.end(), [](int n) { cout << n << " "; });

cout << endl;

}

C:\Temp>cl /EHsc /nologo /W4 meow.cpp > NUL && meow

0 1 2 3 4 5 6 7 8 9

The [] is the lambda-introducer, which tells the compiler that a lambda expression is beginning. The (int n) is the lambda-parameter-declaration, which tells the compiler what the unnamed function object class's function call operator should take. Finally, the { cout << n << " "; } is the compound-statement which serves as the body of the unnamed function object class's function call operator. By default, the unnamed function object class's function call operator returns void.

So, C++0x has mentally translated this into what you'd write in C++98:

C:\Temp>type meow98.cpp

#include <algorithm>

#include <iostream>

#include <ostream>

#include <vector>

using namespace std;

struct LambdaFunctor {

void operator()(int n) const {

cout << n << " ";

}

};

int main() {

vector<int> v;

for (int i = 0; i < 10; ++i) {

v.push_back(i);

}

for_each(v.begin(), v.end(), LambdaFunctor());

cout << endl;

}

C:\Temp>cl /EHsc /nologo /W4 meow98.cpp > NUL && meow98

0 1 2 3 4 5 6 7 8 9

Now I'm going to stop saying things like "the unnamed function object class's function call operator returns void" and start saying "the lambda returns void", but it's important to remember what lambda expressions are doing: defining classes and constructing objects.

Of course, the compound-statement of a lambda can contain multiple statements:

C:\Temp>type multimeow.cpp

#include <algorithm>

#include <iostream>

#include <ostream>

#include <vector>

using namespace std;

int main() {

vector<int> v;

for (int i = 0; i < 10; ++i) {

v.push_back(i);

}

for_each(v.begin(), v.end(), [](int n) {

cout << n;

if (n % 2 == 0) {

cout << " even ";

} else {

cout << " odd ";

}

});

cout << endl;

}

C:\Temp>cl /EHsc /nologo /W4 multimeow.cpp > NUL && multimeow

0 even 1 odd 2 even 3 odd 4 even 5 odd 6 even 7 odd 8 even 9 odd

Now, lambdas don't always have to return void. If a lambda's compound-statement is { return expression; } , then the lambda's return type will be automatically deduced to be the type of expression:

C:\Temp>type cubicmeow.cpp

#include <algorithm>

#include <deque>

#include <iostream>

#include <iterator>

#include <ostream>

#include <vector>

using namespace std;

int main() {

vector<int> v;

for (int i = 0; i < 10; ++i) {

v.push_back(i);

}

deque<int> d;

transform(v.begin(), v.end(), front_inserter(d), [](int n) { return n * n * n; });

for_each(d.begin(), d.end(), [](int n) { cout << n << " "; });

cout << endl;

}

C:\Temp>cl /EHsc /nologo /W4 cubicmeow.cpp > NUL && cubicmeow

729 512 343 216 125 64 27 8 1 0

Here, the type of n * n * n is int, so this lambda's function call operator returns int.

Lambdas with more complicated compound-statements don't get automatically deduced return types. You have to explicitly specify them:

C:\Temp>type returnmeow.cpp

#include <algorithm>

#include <deque>

#include <iostream>

#include <iterator>

#include <ostream>

#include <vector>

using namespace std;

int main() {

vector<int> v;

for (int i = 0; i < 10; ++i) {

v.push_back(i);

}

deque<double> d;

transform(v.begin(), v.end(), front_inserter(d), [](int n) -> double {

if (n % 2 == 0) {

return n * n * n;

} else {

return n / 2.0;

}

});

for_each(d.begin(), d.end(), [](double x) { cout << x << " "; });

cout << endl;

}

C:\Temp>cl /EHsc /nologo /W4 returnmeow.cpp > NUL && returnmeow

4.5 512 3.5 216 2.5 64 1.5 8 0.5 0

The -> double is the optional lambda-return-type-clause. Why doesn't it go on the left, like what programmers have been doing with C functions for longer than I've been alive? Because then the lambda-introducer wouldn't come first, and that's what tells the compiler that a lambda expression is beginning. (Figuring out this kind of stuff is what the Core Working Group excels at; trying to imagine whether a given construct would be parseable within C++ hurts my head.)

If you forget the lambda-return-type-clause, the compiler will complain about each return statement:

C:\Temp>cl /EHsc /nologo /W4 borkedreturnmeow.cpp

borkedreturnmeow.cpp

borkedreturnmeow.cpp(20) : error C3499: a lambda that has been specified to have a void return type cannot return a value

borkedreturnmeow.cpp(22) : error C3499: a lambda that has been specified to have a void return type cannot return a value

All of the lambdas I've presented so far have been stateless: they contain no data members. You can have stateful lambdas too, and this is accomplished through "capturing" local variables. The empty lambda-introducer [] says "I am a stateless lambda". But within the lambda-introducer, you can specify a capture-list:

C:\Temp>type capturekittybyvalue.cpp

#include <algorithm>

#include <iostream>

#include <ostream>

#include <vector>

using namespace std;

int main() {

vector<int> v;

for (int i = 0; i < 10; ++i) {

v.push_back(i);

}

int x = 0;

int y = 0;

// op>>() leaves newlines on the input stream,

// which can be extremely confusing. I recommend

// avoiding it, and instead using non-member

// getline(cin, str) to read whole lines and

// then parse them. But in the interests of

// brevity, I'll use evil op>>():

cout << "Input: ";

cin >> x >> y;

v.erase(remove_if(v.begin(), v.end(), [x, y](int n) { return x < n && n < y; }), v.end());

for_each(v.begin(), v.end(), [](int n) { cout << n << " "; });

cout << endl;

}

C:\Temp>cl /EHsc /nologo /W4 capturekittybyvalue.cpp > NUL && capturekittybyvalue

Input: 4 7

0 1 2 3 4 7 8 9

If you forget the capture-list, the compiler will complain:

C:\Temp>cl /EHsc /nologo /W4 borkedcapturekittybyvalue.cpp

borkedcapturekittybyvalue.cpp

borkedcapturekittybyvalue.cpp(27) : error C3493: 'x' cannot be implicitly captured as no default capture mode has been specified

borkedcapturekittybyvalue.cpp(27) : error C3493: 'y' cannot be implicitly captured as no default capture mode has been specified

(I'll explain default captures soon.)

Remember, the lambda expression is implicitly defining an unnamed function object class. The compound-statement { return x < n && n < y; } serves as the body of the function call operator within that class. Although the compound-statement is lexically within the scope of main(), it is conceptually outside the scope of main(), so you can't use local variables from main() without capturing them within the lambda.

Here's what this is being mentally translated into:

C:\Temp>type capturekittybyvalue98.cpp

#include <algorithm>

#include <iostream>

#include <iterator>

#include <ostream>

#include <vector>

using namespace std;

class LambdaFunctor {

public:

LambdaFunctor(int a, int b) : m_a(a), m_b(b) { }

bool operator()(int n) const { return m_a < n && n < m_b; }

private:

int m_a;

int m_b;

};

int main() {

vector<int> v;

for (int i = 0; i < 10; ++i) {

v.push_back(i);

}

int x = 0;

int y = 0;

cout << "Input: ";

cin >> x >> y; // EVIL!

v.erase(remove_if(v.begin(), v.end(), LambdaFunctor(x, y)), v.end());

copy(v.begin(), v.end(), ostream_iterator<int>(cout, " "));

cout << endl;

}

C:\Temp>cl /EHsc /nologo /W4 capturekittybyvalue98.cpp > NUL && capturekittybyvalue98

Input: 4 7

0 1 2 3 4 7 8 9

Here, you can clearly see that the captures are "by value". Copies of the local variables are being stored within the function object. This allows the function object to outlive the local variables that were captured to create it. However, note that (a) the captured copies can't be modified within the lambda, because by default the function call operator is const, (b) some objects are expensive to copy, and (c) updates to the local variables will not be reflected in the captured copies (these are ordinary value semantics). Soon, I'll explain how to deal with each of these when necessary.

But first, instead of specifying every local variable that you want to capture, you can say "capture everything by value". The syntax for this is the lambda-introducer [=] (the capture-default = is supposed to make you think of assignment or copy-initialization Foo foo = bar; , although the copies are actually made by direct-initialization, like m_a(a) above):

C:\Temp>type defaultcapturekittybyvalue.cpp

#include <algorithm>

#include <iostream>

#include <ostream>

#include <vector>

using namespace std;

int main() {

vector<int> v;

for (int i = 0; i < 10; ++i) {

v.push_back(i);

}

int x = 0;

int y = 0;

cout << "Input: ";

cin >> x >> y; // EVIL!

v.erase(remove_if(v.begin(), v.end(), [=](int n) { return x < n && n < y; }), v.end());

for_each(v.begin(), v.end(), [](int n) { cout << n << " "; });

cout << endl;

}

C:\Temp>cl /EHsc /nologo /W4 defaultcapturekittybyvalue.cpp > NUL && defaultcapturekittybyvalue

Input: 4 7

0 1 2 3 4 7 8 9

When the compiler sees x and y mentioned within the lambda, it captures them from main() by value.

What about (a), modifying the captured copies? By default, a lambda's function call operator is const, but you can make it non-const by saying mutable:

C:\Temp>type capturekittybymutablevalue.cpp

#include <algorithm>

#include <iostream>

#include <ostream>

#include <vector>

using namespace std;

int main() {

vector<int> v;

for (int i = 0; i < 10; ++i) {

v.push_back(i);

}

int x = 1;

int y = 1;

for_each(v.begin(), v.end(), [=](int& r) mutable {

const int old = r;

r *= x * y;

x = y;

y = old;

});

for_each(v.begin(), v.end(), [](int n) { cout << n << " "; });

cout << endl;

cout << x << ", " << y << endl;

}

C:\Temp>cl /EHsc /nologo /W4 capturekittybymutablevalue.cpp > NUL && capturekittybymutablevalue

0 0 0 6 24 60 120 210 336 504

1, 1

This multiplies each element of v by the previous two elements. (I had to think really hard to come up with an example that couldn't be expressed with partial_sum(), which could multiply by all previous elements, or adjacent_difference(), which could multiply by the immediately previous element.) Note that (d) updates to the captured copies are not reflected in the local variables (again, ordinary value semantics).

What if you want to deal with (b), (c), and (d): avoid copies, observe updates to the local variables from within the lambda, and modify the local variables from within the lambda? In that case, you want to capture by reference. The syntax for doing this is the lambda-introducer [&x, &y] (you should think of this as X& x, Y& y ; that is, "reference" and not "address of"):

C:\Temp>type capturekittybyreference.cpp

#include <algorithm>

#include <iostream>

#include <ostream>

#include <vector>

using namespace std;

int main() {

vector<int> v;

for (int i = 0; i < 10; ++i) {

v.push_back(i);

}

int x = 1;

int y = 1;

for_each(v.begin(), v.end(), [&x, &y](int& r) {

const int old = r;

r *= x * y;

x = y;

y = old;

});

for_each(v.begin(), v.end(), [](int n) { cout << n << " "; });

cout << endl;

cout << x << ", " << y << endl;

}

C:\Temp>cl /EHsc /nologo /W4 capturekittybyreference.cpp > NUL && capturekittybyreference

0 0 0 6 24 60 120 210 336 504

8, 9

Note the differences from capturekittybymutablevalue.cpp: (1) the lambda-introducer [&x, &y] , (2) the absence of mutable , and (3) the local variables x and y having values of 8 and 9 at the end, reflecting their modification within the lambda.

This is being mentally translated into:

C:\Temp>type capturekittybyreference98.cpp

#include <algorithm>

#include <iostream>

#include <iterator>

#include <ostream>

#include <vector>

using namespace std;

#pragma warning(push)

#pragma warning(disable: 4512) // assignment operator could not be generated

class LambdaFunctor {

public:

LambdaFunctor(int& a, int& b) : m_a(a), m_b(b) { }

void operator()(int& r) const {

const int old = r;

r *= m_a * m_b;

m_a = m_b;

m_b = old;

}

private:

int& m_a;

int& m_b;

};

#pragma warning(pop)

int main() {

vector<int> v;

for (int i = 0; i < 10; ++i) {

v.push_back(i);

}

int x = 1;

int y = 1;

for_each(v.begin(), v.end(), LambdaFunctor(x, y));

copy(v.begin(), v.end(), ostream_iterator<int>(cout, " "));

cout << endl;

cout << x << ", " << y << endl;

}

C:\Temp>cl /EHsc /nologo /W4 capturekittybyreference98.cpp > NUL && capturekittybyreference98

0 0 0 6 24 60 120 210 336 504

8, 9

(When you use lambdas, the compiler automatically disables warning C4512 for lambda definitions.)

When you capture local variables by reference, the function object stores reference data members to those local variables. This avoids copying, allows the function object to observe updates to those local variables, and allows the function object to modify those local variables through its references. (Note that the function call operator is const, because we didn't say mutable, but that just prevents us from modifying the function object's data members. The data members here are references and can't be modified anyways, but they can be used to modify what they refer to. The constness of the function call operator is shallow, as always.)

Of course, if a lambda function object outlives local variables that it has captured by reference, you get crashtrocity.

Again, you can use default captures; the lambda-introducer [&] says "capture everything by reference".

What if you want to mix value captures and reference captures? You could say [a, b, c, &d, e, &f, g] . But you can also specify a capture-default and then override it for specific local variables. Here's an example, modified from capturekittybymutablevalue.cpp:

C:\Temp>type overridekitty.cpp

#include <algorithm>

#include <iostream>

#include <ostream>

#include <vector>

using namespace std;

int main() {

vector<int> v;

for (int i = 0; i < 10; ++i) {

v.push_back(i);

}

int sum = 0;

int product = 1;

int x = 1;

int y = 1;

for_each(v.begin(), v.end(), [=, &sum, &product](int& r) mutable {

sum += r;

if (r != 0) {

product *= r;

}

const int old = r;

r *= x * y;

x = y;

y = old;

});

for_each(v.begin(), v.end(), [](int n) { cout << n << " "; });

cout << endl;

cout << "sum: " << sum << ", product: " << product << endl;

cout << "x: " << x << ", y: " << y << endl;

}

C:\Temp>cl /EHsc /nologo /W4 overridekitty.cpp && overridekitty

overridekitty.cpp

0 0 0 6 24 60 120 210 336 504

sum: 45, product: 362880

x: 1, y: 1

Here, we want to capture x and y by value (because we want to modify them within the lambda, but not outside), while we want to capture sum and product by reference (because we do want to modify them outside). The opposite lambda-introducer [&, x, y] would produce exactly the same result (capture everything by reference, except x and y by value).

Now, what if you want to do this?

C:\Temp>type memberkitty.cpp

#include <algorithm>

#include <iostream>

#include <ostream>

#include <vector>

using namespace std;

class Kitty {

public:

explicit Kitty(int toys) : m_toys(toys) { }

void meow(const vector<int>& v) const {

for_each(v.begin(), v.end(), [m_toys](int n) {

cout << "If you gave me " << n << " toys, I would have " << n + m_toys << " toys total." << endl;

});

}

private:

int m_toys;

};

int main() {

vector<int> v;

for (int i = 0; i < 3; ++i) {

v.push_back(i);

}

Kitty k(5);

k.meow(v);

}

C:\Temp>cl /EHsc /nologo /W4 memberkitty.cpp

memberkitty.cpp

memberkitty.cpp(12) : error C3480: 'Kitty::m_toys': a lambda capture variable must be from an enclosing function scope

The lambda expression syntax allows you to capture local variables, but data members aren't local variables. By special dispensation, you can also capture this:

C:\Temp>type workingmemberkitty.cpp

#include <algorithm>

#include <iostream>

#include <ostream>

#include <vector>

using namespace std;

class Kitty {

public:

explicit Kitty(int toys) : m_toys(toys) { }

void meow(const vector<int>& v) const {

for_each(v.begin(), v.end(), [this](int n) {

cout << "If you gave me " << n << " toys, I would have " << n + m_toys << " toys total." << endl;

});

}

private:

int m_toys;

};

int main() {

vector<int> v;

for (int i = 0; i < 3; ++i) {

v.push_back(i);

}

Kitty k(5);

k.meow(v);

}

C:\Temp>cl /EHsc /nologo /W4 workingmemberkitty.cpp > NUL && workingmemberkitty

If you gave me 0 toys, I would have 5 toys total.

If you gave me 1 toys, I would have 6 toys total.

If you gave me 2 toys, I would have 7 toys total.

When you've captured this, you can say m_toys, which implicitly means this->m_toys as usual. You could explicitly say this->m_toys too. (Within a lambda expression, you can say this only if you've captured it; you can never get the this pointer for the lambda object itself.)

You can also capture this implicitly:

C:\Temp>type implicitmemberkitty.cpp

#include <algorithm>

#include <iostream>

#include <ostream>

#include <vector>

using namespace std;

class Kitty {

public:

explicit Kitty(int toys) : m_toys(toys) { }

void meow(const vector<int>& v) const {

for_each(v.begin(), v.end(), [=](int n) {

cout << "If you gave me " << n << " toys, I would have " << n + m_toys << " toys total." << endl;

});

}

private:

int m_toys;

};

int main() {

vector<int> v;

for (int i = 0; i < 3; ++i) {

v.push_back(i);

}

Kitty k(5);

k.meow(v);

}

C:\Temp>cl /EHsc /nologo /W4 implicitmemberkitty.cpp > NUL && implicitmemberkitty

If you gave me 0 toys, I would have 5 toys total.

If you gave me 1 toys, I would have 6 toys total.

If you gave me 2 toys, I would have 7 toys total.

You can also say [&] , but it doesn't affect how this is captured (always by value). You can't say [&this] .

If you want a nullary lambda (taking no arguments), you can elide the lambda-parameter-declaration entirely:

C:\Temp>type nullarykitty.cpp

#include <algorithm>

#include <iostream>

#include <iterator>

#include <ostream>

#include <vector>

using namespace std;

int main() {

vector<int> v;

int i = 0;

generate_n(back_inserter(v), 10, [&] { return i++; });

for_each(v.begin(), v.end(), [](int n) { cout << n << " "; });

cout << endl;

cout << "i: " << i << endl;

}

C:\Temp>cl /EHsc /nologo /W4 nullarykitty.cpp > NUL && nullarykitty

0 1 2 3 4 5 6 7 8 9

i: 10

This is two characters shorter than [&]() { return i++; } . Whether eliding the lambda-parameter-declaration is good style is up to you to decide.

For laughs, this means that the following is valid C++0x:

C:\Temp>type nokitty.cpp

int main() {

[](){}();

[]{}();

}

This constructs two do-nothing lambdas (one with a lambda-parameter-declaration, one without) and immediately calls them (that's the final set of empty parentheses).

Note that the optional lambda-parameter-declaration syntactically consists of:

( lambda-parameter-declaration-list_opt ) mutable_opt exception-specification_opt lambda-return-type-clause_opt

So, if you want to say mutable or -> ReturnType, you need empty parentheses between that and the lambda-introducer.

Finally, because lambdas produce ordinary function objects, you can store them in tr1::function:

C:\Temp>type tr1kitty.cpp

#include <algorithm>

#include <functional>

#include <iostream>

#include <ostream>

#include <vector>

using namespace std;

using namespace std::tr1;

void meow(const vector<int>& v, const function<void (int)>& f) {

for_each(v.begin(), v.end(), f);

cout << endl;

}

int main() {

vector<int> v;

for (int i = 0; i < 10; ++i) {

v.push_back(i);

}

meow(v, [](int n) { cout << n << " "; });

meow(v, [](int n) { cout << n * n << " "; });

function<void (int)> g = [](int n) { cout << n * n * n << " "; };

meow(v, g);

}

C:\Temp>cl /EHsc /nologo /W4 tr1kitty.cpp > NUL && tr1kitty

0 1 2 3 4 5 6 7 8 9

0 1 4 9 16 25 36 49 64 81

0 1 8 27 64 125 216 343 512 729

auto

The auto keyword from C++98, where it did absolutely nothing, has been repurposed in C++0x for automatic type deduction. When used in a declaration, it says "make the type of this thing the same as the type of whatever initializes it". Behold:

C:\Temp>type autocat.cpp

#include <iostream>

#include <map>

#include <ostream>

#include <regex>

#include <string>

using namespace std;

using namespace std::tr1;

int main() {

map<string, string> m;

const regex r("(\\w+) (\\w+)");

for (string s; getline(cin, s); ) {

smatch results;

if (regex_match(s, results, r)) {

m[results[1]] = results[2];

}

for (auto i = m.begin(); i != m.end(); ++i) {

cout << i->second << " are " << i->first << endl;

}

C:\Temp>cl /EHsc /nologo /W4 autocat.cpp > NUL && autocat

cute kittens

ugly puppies

evil goblins

kittens are cute

goblins are evil

puppies are ugly

map<string, string>::iterator, your decade-long reign of terror has come to an end!

(Note that m.begin() returns iterator, not const_iterator, because the map is not const. C++0x cbegin() allows you to request a const_iterator from a non-const container.)

lambdas and auto

Earlier, I mentioned storing lambdas in tr1::functions. But you shouldn't do that unless it's necessary, as tr1::function has some overhead. If you want to reuse a lambda, or simply want to give it a name, you can use auto. Here's an example, which also does something neat:

C:\Temp>type autolambdakitty.cpp

#include <algorithm>

#include <iostream>

#include <ostream>

#include <vector>

using namespace std;

template <typename T, typename Predicate> void keep_if(vector<T>& v, Predicate pred) {

auto notpred = [&](const T& t) {

return !pred(t);

};

v.erase(remove_if(v.begin(), v.end(), notpred), v.end());

}

template <typename Container> void print(const Container& c) {

for_each(c.begin(), c.end(), [](const typename Container::value_type& e) { cout << e << " "; });

cout << endl;

}

int main() {

vector<int> a;

for (int i = 0; i < 100; ++i) {

a.push_back(i);

}

vector<int> b;

for (int i = 100; i < 200; ++i) {

b.push_back(i);

}

auto prime = [](const int n) -> bool {

if (n < 2) {

return false;

}

for (int i = 2; i <= n / i; ++i) {

if (n % i == 0) {

return false;

}

return true;

};

keep_if(a, prime);

keep_if(b, prime);

print(a);

print(b);

}

C:\Temp>cl /EHsc /nologo /W4 autolambdakitty.cpp

autolambdakitty.cpp

C:\Temp>autolambdakitty

2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97

101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199

notpred is a negator lambda! Note that we can't use C++98 <functional>'s not1(), as that requires your predicate to derive from unary_function, and lambdas don't.

static_assert

static_assert allows you to trigger compiler errors with custom error messages:

C:\Temp>type staticfluffykitten.cpp

template <int N> struct Kitten {

static_assert(N < 2, "Kitten<N> requires N < 2.");

};

int main() {

Kitten<1> peppermint;

Kitten<3> jazz;

}

C:\Temp>cl /EHsc /nologo /W4 staticfluffykitten.cpp

staticfluffykitten.cpp

staticfluffykitten.cpp(2) : error C2338: Kitten<N> requires N < 2.

staticfluffykitten.cpp(8) : see reference to class template instantiation 'Kitten<N>' being compiled

with

[

N=3

]

Part2: Rvalue References

Today, I'm going to talk about rvalue references, which enable two different things: move semantics and perfect forwarding. This post will be long, because I'm going to explain how rvalue references work in great detail. They're initially very confusing because they distinguish lvalues from rvalues, which very few C++98/03 programmers are extensively familiar with.

Fear not, for using rvalue references is easy, much easier than it initially sounds. Implementing either move semantics or perfect forwarding in your own code boils down to following simple patterns, which I will demonstrate. And it's definitely worth learning how to use rvalue references, as move semantics can produce order of magnitude performance improvements, and perfect forwarding makes writing highly generic code very easy.

lvalues and rvalues in C++98/03

In order to understand rvalue references in C++0x, you must first understand lvalues and rvalues in C++98/03.

The terminology of "lvalues" and "rvalues" is confusing because their history is confusing. (By the way, they're just pronounced as "L values" and "R values", although they're written as single words.) These concepts originally came from C, and then were elaborated upon by C++. To save time, I'll skip over their history, including why they're called "lvalues" and "rvalues", and I'll go directly to how they work in C++98/03. (Okay, it's not a big secret: "L" stands for "left" and "R" stands for "right". But the concepts have evolved since the names were chosen, and the names aren't very accurate anymore. Instead of going through the whole history lesson, you can consider the names to be arbitrary like "up quark" and "down quark", and you won't lose anything.)

C++03 3.10/1 says: "Every expression is either an lvalue or an rvalue." It's important to remember that lvalueness versus rvalueness is a property of expressions, not of objects.

Lvalues name objects that persist beyond a single expression. For example, obj , *ptr , ptr[index] , and ++x are all lvalues.

Rvalues are temporaries that evaporate at the end of the full-expression in which they live ("at the semicolon"). For example, 1729 , x + y , std::string("meow") , and x++ are all rvalues.

Notice the difference between ++x and x++ . If we have int x = 0; then the expression x is an lvalue, as it names a persistent object. The expression ++x is also an lvalue. It modifies and then names the persistent object. However, the expression x++ is an rvalue. It copies the original value of the persistent object, modifies the persistent object, and then returns the copy. This copy is a temporary. Both ++x and x++ increment x, but ++x returns the persistent object itself, while x++ returns a temporary copy. That's why ++x is an lvalue, while x++ is an rvalue. Lvalueness versus rvalueness doesn't care about what an expression does, it cares about what an expression names (something persistent or something temporary).

If you want to build up intuition for this, another way to determine whether an expression is an lvalue is to ask "can I take its address?". If you can, it's an lvalue. If you can't, it's an rvalue. For example, &obj , &*ptr , &ptr[index] , and &++x are all valid (even though some of those expressions are silly), while &1729 , &(x + y) , &std::string("meow") , and &x++ are all invalid. Why does this work? The address-of operator requires that its "operand shall be an lvalue" (C++03 5.3.1/2). Why does it require that? Taking the address of a persistent object is fine, but taking the address of a temporary would be extremely dangerous, because temporaries evaporate quickly.

The preceding examples ignore operator overloading, which is convenient syntax for a function call. "A function call is an lvalue if and only if the result type is a reference." (C++03 5.2.2/10) Therefore, given vector<int> v(10, 1729); , v[0] is an lvalue because operator[]() returns int& (and &v[0] is valid and useful), while given string s("foo"); and string t("bar"); , s + t is an rvalue because operator+() returns string (and &(s + t) is invalid).

Both lvalues and rvalues can be either modifiable (non-const) or non-modifiable (const). Here are examples:

string one("cute");

const string two("fluffy");

string three() { return "kittens"; }

const string four() { return "are an essential part of a healthy diet"; }

one; // modifiable lvalue

two; // const lvalue

three(); // modifiable rvalue

four(); // const rvalue

Type& binds to modifiable lvalues (and can be used to observe and mutate them). It can't bind to const lvalues, as that would violate const correctness. It can't bind to modifiable rvalues, as that would be extremely dangerous. Accidentally modifying temporaries, only to have the temporaries evaporate along with your modifications, would lead to subtle and obnoxious bugs, so C++ rightly prohibits this. (I should mention that VC has an evil extension that allows this, but if you compile with /W4 , it warns when the evil extension is activated. Usually.) And it can't bind to const rvalues, as that would be doubly bad. (Careful readers should note that I'm not talking about template argument deduction here.)

const Type& binds to everything: modifiable lvalues, const lvalues, modifiable rvalues, and const rvalues (and can be used to observe them).

A reference is a name, so a reference bound to an rvalue is itself an lvalue (yes, L). (As only a const reference can be bound to an rvalue, it will be a const lvalue.) This is confusing, and will be an extremely big deal later, so I'll explain further. Given the function void observe(const string& str) , inside observe()'s implementation, str is a const lvalue, and its address can be taken and used before observe() returns. This is true even though observe() can be called with rvalues, such as three() or four() above. observe("purr") can also be called, which constructs a temporary string and binds str to that temporary. The return values of three() and four() don't have names, so they're rvalues, but within observe(), str is a name, so it's an lvalue. As I said above, "lvalueness versus rvalueness is a property of expressions, not of objects". Of course, because str can be bound to a temporary which will evaporate, its address shouldn't be stored anywhere where it could be used after observe() returns.

Have you ever bound an rvalue to a const reference and then taken its address? Yes, you have! This is what happens when you write a copy assignment operator, Foo& operator=(const Foo& other) , with a self-assignment check, if (this != &other) { copy stuff; } return *this; , and you copy assign from a temporary, like Foo make_foo(); Foo f; f = make_foo(); .

At this point, you might ask, "So what's the difference between modifiable rvalues and const rvalues? I can't bind Type& to modifiable rvalues, and I can't assign things (etc.) to modifiable rvalues, so can I really modify them?" This is a very good question! In C++98/03, the answer is that there's a slight difference: non-const member functions can be called on modifiable rvalues. C++ doesn't want you to accidentally modify temporaries, but directly calling a non-const member function on a modifiable rvalue is explicit, so it's allowed. In C++0x, the answer changes dramatically, making move semantics possible.

Congratulations! Now you have what I call "lvalue/rvalue vision", the ability to look at an expression and determine whether it's an lvalue or an rvalue. Combined with your "const vision", you can precisely reason that given void mutate(string& ref) and the definitions above, mutate(one) is valid, while mutate(two), mutate(three()), mutate(four()), and mutate("purr") are invalid, and all of observe(one), observe(two), observe(three()), observe(four()), and observe("purr") are valid. If you're a C++98/03 programmer, you already knew which of these calls were valid and which were invalid; your "gut feeling", if not your compiler, would have told you that mutate(three()) was bogus. Your new lvalue/rvalue vision tells you precisely why (three() is an rvalue, and modifiable references can't be bound to rvalues). Is that useful? To language lawyers, yes, but not really to normal programmers. After all, you've gotten this far without knowing all of this stuff about lvalues and rvalues. But here's the catch: compared to C++98/03, C++0x has vastly more powerful lvalue/rvalue vision (in particular, the ability to look at an expression, determine whether it's a modifiable/const lvalue/rvalue, and do something about it). In order to use C++0x effectively, you need lvalue/rvalue vision too. And now you have it, so we can proceed!

the copying problem

C++98/03 combines insanely powerful abstraction with insanely efficient execution, but it has a problem: it's overly fond of copying. Things with value semantics behave like ints, so copying a thing doesn't modify the source, and the resulting copies are independent. Value semantics are great, except that they tend to lead to unnecessary copies of heavy objects like strings, vectors, and so forth. ("Heavy" means "expensive to copy"; a million-element vector is heavy.) The Return Value Optimization (RVO) and Named Return Value Optimization (NRVO), where copy constructors are elided in certain situations, help to alleviate this problem, but they don't remove all unnecessary copies.

The most unnecessary copies are those where the source is about to be destroyed. Would you photocopy a sheet of paper and then immediately throw away the original, assuming that the original and the photocopy are identical? That would be wasteful; you should keep the original and not bother with the photocopy. Here's what I call "the killer example", derived from one of the Standardization Committee's examples (in N1377). Suppose that you have a bunch of strings, like this:

string s0("my mother told me that");

string s1("cute");

string s2("fluffy");

string s3("kittens");

string s4("are an essential part of a healthy diet");

And that you concatenate them like this:

string dest = s0 + " " + s1 + " " + s2 + " " + s3 + " " + s4;

How efficient is this? (We're not worrying about this specific example, which executes in microseconds; we're worrying about its generalization, which occurs throughout the entire language.)

Each call to operator+() returns a temporary string. There are 8 calls to operator+() , so there are 8 temporary strings. Each one, upon its construction, performs a dynamic memory allocation and copies all of the characters that have been concatenated so far, and later, upon its destruction, performs a dynamic memory deallocation. (If you've heard of the Small String Optimization, which VC performs in order to avoid dynamic memory allocations and deallocations for short strings, it's defeated here by my carefully chosen and sufficiently long s0 , and even if it applied, it couldn't avoid the copying. If you've heard of the Copy-On-Write "optimization", forget about it - it doesn't apply here, and it's a pessimization under multithreading, so Standard Library implementations don't do it anymore.)

In fact, because every concatenation copies all of the characters that have been concatenated so far, this has quadratic complexity in the number of concatenations. Yuck! This is extraordinarily wasteful, which is especially embarrassing for C++. Why is this happening, and what can we do about it?

The problem is that operator+() , which takes two const string& or one const string& and one const char * (there are other overloads, which we aren't using here), can't tell whether it's being fed lvalues versus rvalues, so it always has to create and return a new temporary string . Why do lvalues versus rvalues matter?

When evaluating s0 + " " , it's absolutely necessary to create a new temporary string . s0 is an lvalue, naming a persistent object, so we can't modify it. (Someone would notice!) But when evaluating (s0 + " ") + s1 , we could simply append s1's contents onto our first temporary string, instead of creating a second temporary and throwing the first temporary away. This is the key insight behind move semantics: because s0 + " " is an rvalue, an expression referring to a temporary object, no one else in the entire program can observe that temporary object. If we could detect that expression as being a modifiable rvalue, we could then proceed to modify the temporary object arbitrarily, without anyone else noticing. operator+() isn't "supposed to" modify its arguments, but if they're modifiable rvalues, who cares? In this manner, each call to operator+() can append characters onto a single temporary string . This completely eliminates the unnecessary dynamic memory management and unnecessary copying, leaving us with linear complexity. Yay!

Technically speaking, in C++0x, each call to operator+() still returns a separate temporary string . However, the second temporary string (from evaluating (s0 + " ") + s1 ) is constructed by stealing the memory owned by the first temporary string (from evaluating s0 + " " ) and then appending s1's contents onto that memory (which may trigger an ordinary geometric reallocation). "Stealing" consists of pointer twiddling: the second temporary copies and then nulls out the first temporary's internal pointer. When the first temporary is eventually destroyed ("at the semicolon"), its pointer is null, so its destructor does nothing.

In general, being able to detect modifiable rvalues allows you to engage in "resource pilfering". If the objects referred to by modifiable rvalues own any resources (such as memory), you can steal their resources instead of copying them, since they're going to evaporate anyways. Constructing from or assigning from modifiable rvalues by taking what they own is generically referred to as "moving", and moveable objects have "move semantics".

This is extremely useful in many places, such as vector reallocation. When a vector needs more capacity (e.g. during push_back()) and undergoes reallocation, it needs to copy elements from its old memory block to its new memory block. These copy constructor calls can be expensive. (With a vector<string>, each string needs to be copied, involving dynamic memory allocation.) But wait! The elements in the old memory block are about to be destroyed. So we can move elements over, instead of copying them. In this case, the elements in the old memory block occupy persistent storage, and the expressions used to refer to them, such as old_ptr[index] , are lvalues. During reallocation, we want to refer to the elements in the old memory block with modifiable rvalue expressions. Pretending that they're modifiable rvalues will allow us to move them, eliminating copy constructor calls. (Saying "I want to pretend that this lvalue is a modifiable rvalue" is equivalent to saying "I know that this is an lvalue, referring to a persistent object, but I don't care what happens to the lvalue after this. I'm going to destroy it, or assign to it, or whatever. So if you can steal resources from it, go ahead.")

C++0x's rvalue references enable move semantics by giving us the ability to detect modifiable rvalues and steal from them. Rvalue references also allow us to activate move semantics at will by treating lvalues as modifiable rvalues. Now, let's see how rvalue references work!

rvalue references: initialization

C++0x introduces a new kind of reference, the rvalue reference, with the syntax Type&& and const Type&& . The current C++0x Working Draft, N2798 8.3.2/2, says: "A reference type that is declared using & is called an lvalue reference, and a reference type that is declared using && is called an rvalue reference. Lvalue references and rvalue references are distinct types. Except where explicitly noted, they are semantically equivalent and commonly referred to as references." This means that your intuition for C++98/03 references (now known as lvalue references) translates over to rvalue references; all you have to learn are the differences between them.

(Note: I've settled on pronouncing Type& as "Type ref" and Type&& as "Type ref ref". They're fully known as "lvalue reference to Type" and "rvalue reference to Type", respectively, just like how "const pointer to int" is written as int * const and can be pronounced as "int star const".)

What are the differences? Compared to lvalue references, rvalue references behave differently during initialization and overload resolution. They differ in what they are willing to bind to (i.e. initialization) and what prefers to bind to them (i.e. overload resolution). Let's look at initialization first:

· We've already seen how the modifiable lvalue reference, Type& , is willing to bind to modifiable lvalues, but not to anything else (const lvalues, modifiable rvalues, const rvalues).

· We've already seen how the const lvalue reference, const Type& , is willing to bind to everything.

· The modifiable rvalue reference, Type&& , is willing to bind to modifiable lvalues and modifiable rvalues, but not to const lvalues and const rvalues (which would violate const correctness).

· The const rvalue reference, const Type&& , is willing to bind to everything.

These rules may sound arcane, but they're derived from two simple rules:

· Obey const correctness by preventing modifiable references from binding to const things.

· Avoid accidentally modifying temporaries by preventing modifiable lvalue references from binding to modifiable rvalues.

If you like reading compiler errors instead of reading English, here's a demonstration:

C:\Temp>type initialization.cpp

#include <string>

using namespace std;

string modifiable_rvalue() {

return "cute";

}

const string const_rvalue() {

return "fluffy";

}

int main() {

string modifiable_lvalue("kittens");

const string const_lvalue("hungry hungry zombies");

string& a = modifiable_lvalue; // Line 16

string& b = const_lvalue; // Line 17 - ERROR

string& c = modifiable_rvalue(); // Line 18 - ERROR

string& d = const_rvalue(); // Line 19 - ERROR

const string& e = modifiable_lvalue; // Line 21

const string& f = const_lvalue; // Line 22

const string& g = modifiable_rvalue(); // Line 23

const string& h = const_rvalue(); // Line 24

string&& i = modifiable_lvalue; // Line 26

string&& j = const_lvalue; // Line 27 - ERROR

string&& k = modifiable_rvalue(); // Line 28

string&& l = const_rvalue(); // Line 29 - ERROR

const string&& m = modifiable_lvalue; // Line 31

const string&& n = const_lvalue; // Line 32

const string&& o = modifiable_rvalue(); // Line 33

const string&& p = const_rvalue(); // Line 34

}

C:\Temp>cl /EHsc /nologo /W4 /WX initialization.cpp

initialization.cpp

initialization.cpp(17) : error C2440: 'initializing' : cannot convert from 'const std::string' to 'std::string &'

Conversion loses qualifiers

initialization.cpp(18) : warning C4239: nonstandard extension used : 'initializing' : conversion from 'std::string' to 'std::string &'

A non-const reference may only be bound to an lvalue

initialization.cpp(19) : error C2440: 'initializing' : cannot convert from 'const std::string' to 'std::string &'

Conversion loses qualifiers

initialization.cpp(27) : error C2440: 'initializing' : cannot convert from 'const std::string' to 'std::string &&'

Conversion loses qualifiers

initialization.cpp(29) : error C2440: 'initializing' : cannot convert from 'const std::string' to 'std::string &&'

Conversion loses qualifiers

It's okay for modifiable rvalue references to bind to modifiable rvalues; the whole point is that they can be used to modify temporaries.

Although lvalue references and rvalue references behave similarly during initialization (only lines 18 and 28 above differ), they increasingly diverge during overload resolution.

rvalue references: overload resolution

You're already familiar with how functions can be overloaded on modifiable and const lvalue reference parameters. In C++0x, functions can also be overloaded on modifiable and const rvalue reference parameters. Given all four overloads of a unary function, you shouldn't be surprised to discover that each expression prefers to bind to its corresponding reference:

C:\Temp>type four_overloads.cpp

#include <iostream>

#include <ostream>

#include <string>

using namespace std;

void meow(string& s) {

cout << "meow(string&): " << s << endl;

}

void meow(const string& s) {

cout << "meow(const string&): " << s << endl;

}

void meow(string&& s) {

cout << "meow(string&&): " << s << endl;

}

void meow(const string&& s) {

cout << "meow(const string&&): " << s << endl;

}

string strange() {

return "strange()";

}

const string charm() {

return "charm()";

}

int main() {

string up("up");

const string down("down");

meow(up);

meow(down);

meow(strange());

meow(charm());

}

C:\Temp>cl /EHsc /nologo /W4 four_overloads.cpp

four_overloads.cpp

C:\Temp>four_overloads

meow(string&): up

meow(const string&): down

meow(string&&): strange()

meow(const string&&): charm()

In practice, overloading on Type& , const Type& , Type&& , and const Type&& is not very useful. A far more interesting overload set is const Type& and Type&& :

C:\Temp>type two_overloads.cpp

#include <iostream>

#include <ostream>

#include <string>

using namespace std;

void purr(const string& s) {

cout << "purr(const string&): " << s << endl;

}

void purr(string&& s) {

cout << "purr(string&&): " << s << endl;

}

string strange() {

return "strange()";

}

const string charm() {

return "charm()";

}

int main() {

string up("up");

const string down("down");

purr(up);

purr(down);

purr(strange());

purr(charm());

}

C:\Temp>cl /EHsc /nologo /W4 two_overloads.cpp

two_overloads.cpp

C:\Temp>two_overloads

purr(const string&): up

purr(const string&): down

purr(string&&): strange()

purr(const string&): charm()

How does this work? Here are the rules:

· The initialization rules have veto power.

· Lvalues strongly prefer binding to lvalue references, and rvalues strongly prefer binding to rvalue references.

· Modifiable expressions weakly prefer binding to modifiable references.

(By "veto", I mean that candidate functions which would involve forbidden bindings of expressions to references are immediately deemed to be "non-viable" and are removed from further consideration.) Let's walk through the process of applying the rules.

· For purr(up) , the initialization rules veto neither purr(const string&) nor purr(string&&) . up is an lvalue, so it strongly prefers binding to the lvalue reference purr(const string&) . up is modifiable, so it weakly prefers binding to the modifiable reference purr(string&&) . The strongly preferred purr(const string&) wins.

· For purr(down) , the initialization rules veto purr(string&&) due to const correctness, so purr(const string&) wins by default.

· For purr(strange()) , the initialization rules veto neither purr(const string&) nor purr(string&&) . strange() is an rvalue, so it strongly prefers binding to the rvalue reference purr(string&&) . strange() is modifiable, so it weakly prefers binding to the modifiable reference purr(string&&) . The doubly preferred purr(string&&) wins.

· For purr(charm()) , the initialization rules veto purr(string&&) due to const correctness, so purr(const string&) wins by default.

The important thing to notice is that when you overload on const Type& and Type&& , modifiable rvalues bind to Type&& , while everything else binds to const Type& . Therefore, this is the overload set for move semantics.

Important note: functions returning by value should return Type (like strange()) instead of const Type (like charm()). The latter buys you virtually nothing (forbidding non-const member function calls) and prevents the move semantics optimization.

move semantics: the pattern

Here's a simple class, remote_integer, that stores a pointer to a dynamically allocated int . (This is "remote ownership".) Its default constructor, unary constructor, copy constructor, copy assignment operator, and destructor should all look very familiar to you. I've additionally given it a move constructor and move assignment operator. They're guarded by #ifdef MOVABLE so that I can demonstrate what happens with and without them; real code won't do this.

C:\Temp>type remote.cpp

#include <stddef.h>

#include <iostream>

#include <ostream>

using namespace std;

class remote_integer {

public:

remote_integer() {

cout << "Default constructor." << endl;

m_p = NULL;

}

explicit remote_integer(const int n) {

cout << "Unary constructor." << endl;

m_p = new int(n);

}

remote_integer(const remote_integer& other) {

cout << "Copy constructor." << endl;

if (other.m_p) {

m_p = new int(*other.m_p);

} else {

m_p = NULL;

}

#ifdef MOVABLE

remote_integer(remote_integer&& other) {

cout << "MOVE CONSTRUCTOR." << endl;

m_p = other.m_p;

other.m_p = NULL;

}

#endif // #ifdef MOVABLE

remote_integer& operator=(const remote_integer& other) {

cout << "Copy assignment operator." << endl;

if (this != &other) {

delete m_p;

if (other.m_p) {

m_p = new int(*other.m_p);

} else {

m_p = NULL;

}

return *this;

}

#ifdef MOVABLE

remote_integer& operator=(remote_integer&& other) {

cout << "MOVE ASSIGNMENT OPERATOR." << endl;

if (this != &other) {

delete m_p;

m_p = other.m_p;

other.m_p = NULL;

}

return *this;

}

#endif // #ifdef MOVABLE

~remote_integer() {

cout << "Destructor." << endl;

delete m_p;

}

int get() const {

return m_p ? *m_p : 0;

}

private:

int * m_p;

};

remote_integer square(const remote_integer& r) {

const int i = r.get();

return remote_integer(i * i);

}

int main() {

remote_integer a(8);

cout << a.get() << endl;

remote_integer b(10);

cout << b.get() << endl;

b = square(a);

cout << b.get() << endl;

}

C:\Temp>cl /EHsc /nologo /W4 remote.cpp

remote.cpp

C:\Temp>remote

Unary constructor.

Copy assignment operator.

Destructor.

C:\Temp>cl /EHsc /nologo /W4 /DMOVABLE remote.cpp

remote.cpp

C:\Temp>remote

Unary constructor.

MOVE ASSIGNMENT OPERATOR.

Destructor.

There are several things to notice here.

· The copy and move constructors are overloaded, and the copy and move assignment operators are overloaded. We've already seen what happens to functions overloaded on const Type& and Type&& . This is what allows b = square(a); to automatically select the move assignment operator when it's available.

· Instead of dynamically allocating memory, the move constructor and move assignment operator simply steal it from other . When stealing, we copy other's pointer and then null it out. When other is destroyed, its destructor will do nothing.

· Both the copy and move assignment operators need self-assignment checks. It's well-known why copy assignment operators need self-assignment checks. This is because plain old data types like ints can be assigned to themselves harmlessly (e.g. with x = x;), so user-defined data types should also be harmlessly self-assignable. Self-assignment virtually never happens in handwritten code, but it can easily happen inside algorithms like std::sort() . In C++0x, algorithms like std::sort() can move elements around instead of copying them. The same potential for self-assignment exists here.

At this point, you may be wondering how this interacts with automatically generated ("implicitly declared" in Standardese) constructors and assignment operators.

· Move constructors and move assignment operators are never implicitly declared.

· The implicit declaration of a default constructor is inhibited by any user-declared constructors, including copy constructors and move constructors.

· The implicit declaration of a copy constructor is inhibited by a user-declared copy constructor, but not a user-declared move constructor.

· The implicit declaration of a copy assignment operator is inhibited by a user-declared copy assignment operator, but not a user-declared move assignment operator.

Basically, the automatic generation rules don't interact with move semantics, except that declaring a move constructor, like declaring any constructor, inhibits the implicitly declared default constructor.

move semantics: moving from lvalues

Now, what if you like to write your copy constructors in terms of your copy assignment operators? You might attempt to write your move constructors in terms of your move assignment operators. This is possible, but you have to be careful. Here's the wrong way to do it:

C:\Temp>type unified_wrong.cpp

#include <stddef.h>

#include <iostream>

#include <ostream>

using namespace std;

class remote_integer {

public:

remote_integer() {

cout << "Default constructor." << endl;

m_p = NULL;

}

explicit remote_integer(const int n) {

cout << "Unary constructor." << endl;

m_p = new int(n);

}

remote_integer(const remote_integer& other) {

cout << "Copy constructor." << endl;

m_p = NULL;

*this = other;

}

#ifdef MOVABLE

remote_integer(remote_integer&& other) {

cout << "MOVE CONSTRUCTOR." << endl;

m_p = NULL;

*this = other; // WRONG

}

#endif // #ifdef MOVABLE

remote_integer& operator=(const remote_integer& other) {

cout << "Copy assignment operator." << endl;

if (this != &other) {

delete m_p;

if (other.m_p) {

m_p = new int(*other.m_p);

} else {

m_p = NULL;

}

return *this;

}

#ifdef MOVABLE

remote_integer& operator=(remote_integer&& other) {

cout << "MOVE ASSIGNMENT OPERATOR." << endl;

if (this != &other) {

delete m_p;

m_p = other.m_p;

other.m_p = NULL;

}

return *this;

}

#endif // #ifdef MOVABLE

~remote_integer() {

cout << "Destructor." << endl;

delete m_p;

}

int get() const {

return m_p ? *m_p : 0;

}

private:

int * m_p;

};

remote_integer frumple(const int n) {

if (n == 1729) {

return remote_integer(1729);

}

remote_integer ret(n * n);

return ret;

}

int main() {

remote_integer x = frumple(5);

cout << x.get() << endl;

remote_integer y = frumple(1729);

cout << y.get() << endl;

}

C:\Temp>cl /EHsc /nologo /W4 /O2 unified_wrong.cpp

unified_wrong.cpp

C:\Temp>unified_wrong

Unary constructor.

Copy constructor.

Copy assignment operator.

Destructor.

Unary constructor.

1729

Destructor.

C:\Temp>cl /EHsc /nologo /W4 /O2 /DMOVABLE unified_wrong.cpp

unified_wrong.cpp

C:\Temp>unified_wrong

Unary constructor.

MOVE CONSTRUCTOR.

Copy assignment operator.

Destructor.

Unary constructor.

1729

Destructor.

(The compiler is performing the RVO here, but not the NRVO. As I mentioned earlier, some copy constructor calls are elided by the RVO and NRVO, but the compiler isn't always able to apply them. Move constructors optimize the remaining cases.)

The line marked WRONG inside the move constructor is calling the copy assignment operator! This compiles and runs, but it defeats the purpose of the move constructor.

What happened? Remember from C++98/03 that named lvalue references are lvalues (if you say int& r = *p; then r is an lvalue) and unnamed lvalue references are also lvalues (given vector<int> v(10, 1729), calling v[0] returns int& , an unnamed lvalue reference whose address you can take). Rvalue references behave differently:

· Named rvalue references are lvalues.

· Unnamed rvalue references are rvalues.

A named rvalue reference is an lvalue because it can be repeatedly mentioned, with multiple operations performed on it. If instead it were an rvalue, then the first operation performed could steal from it, affecting subsequent operations. Stealing is supposed to be unobservable, so this is forbidden. On the other hand, unnamed rvalue references can't be repeatedly mentioned, so they can preserve their rvalueness.

If you're really intent on implementing your move constructors in terms of your move assignment operators, you'll need the ability to move from lvalues by treating them as rvalues. This is powered by std::move() from C++0x <utility>, which will be in VC10 (in fact, it's already in my development build), but because it's not in the VC10 CTP, I'll show you how to write it from scratch:

C:\Temp>type unified_right.cpp

#include <stddef.h>

#include <iostream>

#include <ostream>

using namespace std;

template <typename T> struct RemoveReference {

typedef T type;

};

template <typename T> struct RemoveReference<T&> {

typedef T type;

};

template <typename T> struct RemoveReference<T&&> {

typedef T type;

};

template <typename T> typename RemoveReference<T>::type&& Move(T&& t) {

return t;

}

class remote_integer {

public:

remote_integer() {

cout << "Default constructor." << endl;

m_p = NULL;

}

explicit remote_integer(const int n) {

cout << "Unary constructor." << endl;

m_p = new int(n);

}

remote_integer(const remote_integer& other) {

cout << "Copy constructor." << endl;

m_p = NULL;

*this = other;

}

#ifdef MOVABLE

remote_integer(remote_integer&& other) {

cout << "MOVE CONSTRUCTOR." << endl;

m_p = NULL;

*this = Move(other); // RIGHT

}

#endif // #ifdef MOVABLE

remote_integer& operator=(const remote_integer& other) {

cout << "Copy assignment operator." << endl;

if (this != &other) {

delete m_p;

if (other.m_p) {

m_p = new int(*other.m_p);

} else {

m_p = NULL;

}

return *this;

}

#ifdef MOVABLE

remote_integer& operator=(remote_integer&& other) {

cout << "MOVE ASSIGNMENT OPERATOR." << endl;

if (this != &other) {

delete m_p;

m_p = other.m_p;

other.m_p = NULL;

}

return *this;

}

#endif // #ifdef MOVABLE

~remote_integer() {

cout << "Destructor." << endl;

delete m_p;

}

int get() const {

return m_p ? *m_p : 0;

}

private:

int * m_p;

};

remote_integer frumple(const int n) {

if (n == 1729) {

return remote_integer(1729);

}

remote_integer ret(n * n);

return ret;

}

int main() {

remote_integer x = frumple(5);

cout << x.get() << endl;

remote_integer y = frumple(1729);

cout << y.get() << endl;

}

C:\Temp>cl /EHsc /nologo /W4 /O2 /DMOVABLE unified_right.cpp

unified_right.cpp

C:\Temp>unified_right

Unary constructor.

MOVE CONSTRUCTOR.

MOVE ASSIGNMENT OPERATOR.

Destructor.

Unary constructor.

1729

Destructor.

(I'll refer to std::move() and my Move() interchangeably, since they're implemented identically.) How does std::move() work? For the moment, I'm going to say that "magic is involved". (There's a full explanation below; it's not complicated, but it involves template argument deduction and reference collapsing, which we'll encounter while looking at perfect forwarding.) I can skip over the magic with concrete examples. Given an lvalue of type string, like up from the overload resolution examples above, std::move(up) calls string&& std::move(string&) . This returns an unnamed rvalue reference, which is an rvalue. Given an rvalue of type string, like strange() from above, std::move(strange()) calls string&& std::move(string&&) . Again, this returns an unnamed rvalue reference, which is an rvalue.

std::move() is useful in other places than implementing move constructors in terms of move assignment operators. Whenever you reach a point where you can say "I have an lvalue here, and its value is going to cease to matter" (e.g. because it's going to be destroyed or assigned to), you can write std::move(your lvalue expression) in order to activate move semantics.

move semantics: movable members

C++0x's Standard classes (e.g. vector, string, regex) have move constructors and move assignment operators, and we've seen how to implement them in our own classes that manually manage resources (e.g. remote_integer). But what about classes containing movable data members (e.g. vector, string, regex, remote_integer)? The compiler won't automatically generate move constructors and move assignment operators for us. So, we'll need to write them ourselves. Fortunately, with std::move() , this is extremely easy:

C:\Temp>type point.cpp

#include <stddef.h>

#include <iostream>

#include <ostream>

using namespace std;

template <typename T> struct RemoveReference {

typedef T type;

};

template <typename T> struct RemoveReference<T&> {

typedef T type;

};

template <typename T> struct RemoveReference<T&&> {

typedef T type;

};

template <typename T> typename RemoveReference<T>::type&& Move(T&& t) {

return t;

}

class remote_integer {

public:

remote_integer() {

cout << "Default constructor." << endl;

m_p = NULL;

}

explicit remote_integer(const int n) {

cout << "Unary constructor." << endl;

m_p = new int(n);

}

remote_integer(const remote_integer& other) {

cout << "Copy constructor." << endl;

if (other.m_p) {

m_p = new int(*other.m_p);

} else {

m_p = NULL;

}

remote_integer(remote_integer&& other) {

cout << "MOVE CONSTRUCTOR." << endl;

m_p = other.m_p;

other.m_p = NULL;

}

remote_integer& operator=(const remote_integer& other) {

cout << "Copy assignment operator." << endl;

if (this != &other) {

delete m_p;

if (other.m_p) {

m_p = new int(*other.m_p);

} else {

m_p = NULL;

}

return *this;

}

remote_integer& operator=(remote_integer&& other) {

cout << "MOVE ASSIGNMENT OPERATOR." << endl;

if (this != &other) {

delete m_p;

m_p = other.m_p;

other.m_p = NULL;

}

return *this;

}

~remote_integer() {

cout << "Destructor." << endl;

delete m_p;

}

int get() const {

return m_p ? *m_p : 0;

}

private:

int * m_p;

};

class remote_point {

public:

remote_point(const int x_arg, const int y_arg)

: m_x(x_arg), m_y(y_arg) { }

remote_point(remote_point&& other)

: m_x(Move(other.m_x)),

m_y(Move(other.m_y)) { }

remote_point& operator=(remote_point&& other) {

m_x = Move(other.m_x);

m_y = Move(other.m_y);

return *this;

}

int x() const { return m_x.get(); }

int y() const { return m_y.get(); }

private:

remote_integer m_x;

remote_integer m_y;

};

remote_point five_by_five() {

return remote_point(5, 5);

}

remote_point taxicab(const int n) {

if (n == 0) {

return remote_point(1, 1728);

}

remote_point ret(729, 1000);

return ret;

}

int main() {

remote_point p = taxicab(43112609);

cout << "(" << p.x() << ", " << p.y() << ")" << endl;

p = five_by_five();

cout << "(" << p.x() << ", " << p.y() << ")" << endl;

}

C:\Temp>cl /EHsc /nologo /W4 /O2 point.cpp

point.cpp

C:\Temp>point

Unary constructor.

MOVE CONSTRUCTOR.

Destructor.

(729, 1000)

Unary constructor.

MOVE ASSIGNMENT OPERATOR.

Destructor.

(5, 5)

Destructor.

As you can see, memberwise moves are trivial to write. Note that remote_point's move assignment operator doesn't need to check for self-assignment because remote_integer already does. Also note that remote_point's implicitly declared copy constructor, copy assignment operator, and destructor do the right things.

By now, you should be exhaustively familiar with move semantics. (But hopefully not exhausted!) As a test of your newly acquired incredible powers, solving the original example of the copying problem with operator+() (applied to remote_integer instead of string) is an exercise left to the reader.

A final reminder: whenever possible, you should implement move constructors and move assignment operators for your copyable classes, because the compiler won't do it for you. Not only will ordinary use of those classes automatically pick up move semantics, but STL containers and algorithms will also take advantage of move semantics, replacing expensive copies with cheap moves.

the forwarding problem

C++98/03's rules for lvalues, rvalues, references, and templates seemed perfectly fine until programmers tried to write highly generic code. Suppose you're writing a completely generic function outer() whose purpose in life is to take an arbitrary number of arbitrary arguments and pass them ("forward" them) to an arbitrary function inner() . There are many good examples of this. The factory function make_shared<T>(args) forwards args to T's constructor and returns shared_ptr<T> . (This can store the T object and its reference count control block in a single dynamic memory allocation, achieving efficiency equal to that of intrusive reference counting.) Wrappers like function<Ret (Args)> forward arguments to their stored functors, and so forth. In this post, our only interest is outer() forwarding arguments to inner() . Determining outer()'s return type is a separate matter (some cases are easy, e.g. make_shared<T>(args) always returns shared_ptr<T>, but the C++0x feature decltype is required to solve the fully general case).

The zero-argument case solves itself, so what about the one-argument case? Let's try to write outer() :

template <typename T> void outer(T& t) {

inner(t);

}

The problem with this outer() is that it can't be called with modifiable rvalues. If inner() takes const int& , then inner(5) will compile. But outer(5) won't compile; T will be deduced to be int, and int& won't bind to 5 .

Okay, so let's try this:

template <typename T> void outer(const T& t) {

inner(t);

}

If inner() takes int& , then this will violate const correctness, so it won't compile.

Now, you can overload outer() on both T& and const T& , which actually works. Then you can call outer() exactly as if it were inner() .

Unfortunately, this doesn't scale to the multi-argument case. That would involve overloading on T1& and const T1& , T2& and const T2& , etc. for every argument, producing an exponential number of overloads. (VC9 SP1's tr1::bind() is desperate enough to do this for the first 5 arguments, involving 63 overloads. Otherwise, it would be hard to explain to users why they couldn't call bound functors with rvalues like 1729 without going through this whole explanation. Stamping out these overloads requires disgusting preprocessor machinery that you really don't want to know about.)

The forwarding problem is severe and fundamentally unsolvable in C++98/03 (without resorting to that disgusting preprocessor machinery that significantly slows down compilation and makes code virtually unreadable). However, rvalue references solve the forwarding problem in an elegant manner.

(I explained the initialization and overload resolution rules before demonstrating the move semantics pattern, but now I'm going to demonstrate the perfect forwarding pattern before explaining the template argument deduction and reference collapsing rules. It'll make more sense this way.)

perfect forwarding: the pattern

Perfect forwarding allows you to write a single function template that takes N arbitrary arguments and forwards them, transparently, to an arbitrary function. Their modifiable/const lvalue/rvalue nature will be preserved, permitting outer() to be used exactly like inner() , and cooperating with move semantics as a bonus. (C++0x variadic templates solve the "arbitrary number" part; we'll be considering any given N here.) This is easy, although it looks like magic at first:

C:\Temp>type perfect.cpp

#include <iostream>

#include <ostream>

using namespace std;

template <typename T> struct Identity {

typedef T type;

};

template <typename T> T&& Forward(typename Identity<T>::type&& t) {

return t;

}

void inner(int&, int&) {

cout << "inner(int&, int&)" << endl;

}

void inner(int&, const int&) {

cout << "inner(int&, const int&)" << endl;

}

void inner(const int&, int&) {

cout << "inner(const int&, int&)" << endl;

}

void inner(const int&, const int&) {

cout << "inner(const int&, const int&)" << endl;

}

template <typename T1, typename T2> void outer(T1&& t1, T2&& t2) {

inner(Forward<T1>(t1), Forward<T2>(t2));

}

int main() {

int a = 1;

const int b = 2;

cout << "Directly calling inner()." << endl;

inner(a, a);

inner(b, b);

inner(3, 3);

inner(a, b);

inner(b, a);

inner(a, 3);

inner(3, a);

inner(b, 3);

inner(3, b);

cout << endl << "Calling outer()." << endl;

outer(a, a);

outer(b, b);

outer(3, 3);

outer(a, b);

outer(b, a);

outer(a, 3);

outer(3, a);

outer(b, 3);

outer(3, b);

}

C:\Temp>cl /EHsc /nologo /W4 perfect.cpp

perfect.cpp

C:\Temp>perfect

Directly calling inner().

inner(int&, int&)

inner(const int&, const int&)

inner(int&, const int&)

inner(const int&, int&)

inner(int&, const int&)

inner(const int&, int&)

inner(const int&, const int&)

Calling outer().

inner(int&, int&)

inner(const int&, const int&)

inner(int&, const int&)

inner(const int&, int&)

inner(int&, const int&)

inner(const int&, int&)

inner(const int&, const int&)

Two lines! Perfect forwarding takes two lines! That's neat.

This demonstrates that outer() forwards t1 and t2 to inner() transparently; inner() observes lvalueness/rvalueness and constness as if it had been directly called.

Like std::move(), std::identity and std::forward() are defined by C++0x <utility> (and will be in VC10, but aren't in the CTP); I'm demonstrating how they're implemented. (Again, I'll refer to std::identity and my Identity, and std::forward() and my Forward(), interchangeably, since they're implemented identically.)

Now, let's figure out how the magic works. It's powered by template argument deduction and reference collapsing.

rvalue references: template argument deduction and reference collapsing

Rvalue references and templates interact in a special way. This demonstrates what happens:

C:\Temp>type collapse.cpp

#include <iostream>

#include <ostream>

#include <string>

using namespace std;

template <typename T> struct Name;

template <> struct Name<string> {

static const char * get() {

return "string";

}

};

template <> struct Name<const string> {

static const char * get() {

return "const string";

}

};

template <> struct Name<string&> {

static const char * get() {

return "string&";

}

};

template <> struct Name<const string&> {

static const char * get() {

return "const string&";

}

};

template <> struct Name<string&&> {

static const char * get() {

return "string&&";

}

};

template <> struct Name<const string&&> {

static const char * get() {

return "const string&&";

}

};

template <typename T> void quark(T&& t) {

cout << "t: " << t << endl;

cout << "T: " << Name<T>::get() << endl;

cout << "T&&: " << Name<T&&>::get() << endl;

cout << endl;

}

string strange() {

return "strange()";

}

const string charm() {

return "charm()";

}

int main() {

string up("up");

const string down("down");

quark(up);

quark(down);

quark(strange());

quark(charm());

}

C:\Temp>cl /EHsc /nologo /W4 collapse.cpp

collapse.cpp

C:\Temp>collapse

t: up

T: string&

T&&: string&

t: down

T: const string&

T&&: const string&

t: strange()

T: string

T&&: string&&

t: charm()

T: const string

T&&: const string&&

The explicit specializations of Name allow us to print out types.

When we call quark(up), template argument deduction is performed. quark() is a function template with a template parameter T, but we haven't provided an explicit template argument (which would look like quark<X>(up)). Instead, a template argument can be deduced by comparing the function parameter type T&& with the function argument type (an lvalue of type string).

C++0x transforms both the function parameter type and the function argument type before matching them together.

First, it transforms the function argument type. A special rule is activated (N2798 14.8.2.1 [temp.deduct.call]/3): when the function parameter type is of the form T&& where T is a template parameter, and the function argument is an lvalue of type A, the type A& is used for template argument deduction. (This special rule doesn't apply to function parameter types of the form T& or const T& , which behave as they did in C++98/03, nor does it apply to const T&& .) In the case of quark(up), this means that we transform string into string& .

Then, it transforms the function parameter type. Both C++98/03 and C++0x discard references (C++0x discards both lvalue references and rvalue references). In all four cases, this means that we transform T&& into T .

Therefore, we deduce T to be the transformed function argument type. That's why quark(up) prints "T: string&" and quark(down) prints "T: const string&"; up and down are lvalues, so they activate the special rule. strange() and charm() are rvalues, so they use the ordinary rules, which is why quark(strange()) prints "T: string" and quark(charm()) prints "T: const string".

Substitution is performed after template argument deduction. Each occurrence of the template parameter T is replaced by the deduced template argument. In quark(strange()), T is string, so T&& is string&& . Similarly, in quark(charm()), T is const string, so T&& is const string&& . However, quark(up) and quark(down) activate another special rule.

In quark(up) , T is string& . Substituting that into T&& produces string& && . References to references are collapsed in C++0x, and the reference collapsing rule is that "lvalue references are infectious". X& & , X& && , and X&& & collapse to X& . Only X&& && collapses to X&& . So, string& && collapses to string& . In templates, things that look like rvalue references aren't necessarily so. quark(up) instantiates quark<string&>(). Within this instantiation, T&& is string& . We've observed this with Name<T&&>::get(). Similarly, when quark(down) instantiates quark<const string&>(), T&& is const string& . In C++98/03, you're probably used to constness hiding inside template parameters (a function template taking T& can be called with a const Foo object; what looks like T& will be const Foo&). In C++0x, lvalueness can hide inside template parameters.

Okay, so what do these two special rules buy us? Within quark(), the type T&& has the same lvalueness/rvalueness and constness as quark()'s function argument. This is how rvalue references preserve lvalueness/rvalueness and constness for perfect forwarding.

perfect forwarding: how std::forward() and std::identity work

Let's look at outer() again:

template <typename T1, typename T2> void outer(T1&& t1, T2&& t2) {

inner(Forward<T1>(t1), Forward<T2>(t2));

}

Now we understand why outer() takes T1&& and T2&& . These types preserve information about outer()'s arguments. But why does it call Forward<T1>() and Forward<T2>() ? Recall that both named lvalue references and named rvalue references are lvalues. If outer() called inner(t1, t2), then inner() would always observe t1 and t2 as being lvalues, breaking perfect forwarding.

Fortunately, unnamed lvalue references are lvalues, and unnamed rvalue references are rvalues. So, in order to forward t1 and t2 to inner(), we need to pass them through helper functions that preserve their types but remove their names. This is what std::forward() does:

template <typename T> struct Identity {

typedef T type;

};

template <typename T> T&& Forward(typename Identity<T>::type&& t) {

return t;

}

When we call Forward<T1>(t1), Identity doesn't modify T1 (we'll soon see what it's doing). So Forward<T1>() takes T1&& and returns T1&& . This leaves t1's type unchanged (whatever it is; we've seen that it can be string& , const string& , string&& , or const string&&) but removes its name. inner() will observe Forward<T1>(t1) as having the same type, lvalueness/rvalueness, and constness as whatever was passed as outer()'s first argument. That's how perfect forwarding works!

You might wonder what would happen if you accidentally wrote Forward<T1&&>(t1) . (This mistake is slightly tempting, since outer() takes T1&& t1 .) Fortunately, nothing bad would happen. Forward<T1&&>() would take and return T1&& && . This collapses to T1&& . Therefore, Forward<T1>(t1) and Forward<T1&&>(t1) are identical, but the former is shorter, so it should be preferred.

What is Identity doing? Why wouldn't this work:

template <typename T> T&& Forward(T&& t) { // BROKEN

return t;

}

If Forward() were written like that, it could be called without explicit template arguments. Template argument deduction would kick in, and we've seen what happens to T&& - it becomes an lvalue reference when its function argument is an lvalue. And the whole problem that we're trying to solve with Forward() is that within outer(), the named t1 and t2 are lvalues even when their types T1&& and T2&& are rvalue references. With the BROKEN implementation, Forward<T1>(t1) would still work, but Forward(t1) would compile (so tempting!) and do the wrong thing, acting just like plain t1 . That would be a continual source of pain, so Identity is used to disable template argument deduction. Experienced template programmers should be familiar with this, as it works identically in C++98/03 and C++0x: the double colon in typename Identity<T>::type is a sheet of lead, and template argument deduction can't see through to the left of it. (Explaining that is another story.)

move semantics: how std::move() works

Now that we've learned the special rules for template argument deduction and reference collapsing, let's look at std::move() again:

template <typename T> struct RemoveReference {

typedef T type;

};

template <typename T> struct RemoveReference<T&> {

typedef T type;

};

template <typename T> struct RemoveReference<T&&> {

typedef T type;

};

template <typename T> typename RemoveReference<T>::type&& Move(T&& t) {

return t;

}

The RemoveReference machinery exactly replicates std::remove_reference from C++0x <type_traits>. For example, RemoveReference<string>::type , RemoveReference<string&>::type , and RemoveReference<string&&>::type are all string .

Similarly, the Move() machinery exactly replicates std::move() from C++0x <utility>.

· When called with an lvalue of type string, T is deduced to be string& , so Move() takes string& (after collapsing) and returns string&& (after RemoveReference).

· When called with an lvalue of type const string, T is deduced to be const string& , so Move() takes const string& (after collapsing) and returns const string&& (after RemoveReference).

· When called with an rvalue of type string, T is deduced to be string , so Move() takes string&& and returns string&& .

· When called with an rvalue of type const string, T is deduced to be const string , so Move() takes const string&& and returns const string&& .

This is how Move() preserves the type and constness of its argument, but converts lvalues into rvalues.

the past

To learn more about rvalue references, you can read their proposal papers. Note that things have changed since these proposals were written; rvalue references have been integrated into the C++0x Working Paper, where they're continuing to receive wording tweaks. Some parts of the proposals are incorrect, outdated, or discussing alternatives that weren't adopted. Still, they're very informative.

N1377, N1385, and N1690 are the main proposal papers. N2118 contains the final revision of the proposed wording before it was integrated into the Working Paper. N1784, N1821, N2377, and N2439 trace the evolution of "Extending Move Semantics To *this", which has been integrated into C++0x but not yet implemented in VC10.

the future

N2812 "A Safety Problem With Rvalue References (and what to do about it)" proposes a change to the initialization rules, forbidding rvalue references from binding to lvalues. It turns out that this won't affect the move semantics and perfect forwarding patterns, so it won't invalidate the new techniques that you just learned (but it will change how std::move() and std::forward() are implemented).

Stephan T. Lavavej

Visual C++ Libraries Developer

Part3: decltype

Today, I'm going to talk about decltype, which allows perfect forwarding functions to have arbitrary return types. It's of interest to people who are writing highly generic code.

the return type problem

C++98/03 has an interesting blind spot - given an expression like x * y, where x and y have arbitrary types, there's no way to say "the type of x * y". If x is of type Watts and y is of type Seconds, then x * y might be of type Joules. Given print(const T& t), you can call print(x * y), and T will be deduced to be Joules, but this doesn't work in reverse: when writing multiply(const A& a, const B& b), you can't name its return type while preserving full generality. Even though when multiply<A, B>() is instantiated, the compiler knows the type of x * y, that information is unavailable to you here. The C++0x keyword decltype removes this blind spot, allowing you to say "multiply() returns the type of x * y". (decltype is an abbreviation of "declared type"; I pronounce it as rhyming with "speckle type".)

decltype: the pattern

Here's how to write a completely generic functor that wraps operator+(). This Plus functor is not a template, but it has a templated function call operator that takes two arguments of arbitrary (and possibly different) types, adds them together, and returns the result, which can be of arbitrary (and possibly different from both of the arguments) type.

C:\Temp>type plus.cpp

#include <algorithm>

#include <iostream>

#include <iterator>

#include <ostream>

#include <string>

#include <utility>

#include <vector>

using namespace std;

struct Plus {

template <typename T, typename U>

auto operator()(T&& t, U&& u) const

-> decltype(forward<T>(t) + forward(u)) {

return forward<T>(t) + forward(u);

}

};

int main() {

vector<int> i;

i.push_back(1);

i.push_back(2);

i.push_back(3);

vector<int> j;

j.push_back(40);

j.push_back(50);

j.push_back(60);

vector<int> k;

vector<string> s;

s.push_back("cut");

s.push_back("flu");

s.push_back("kit");

vector<string> t;

t.push_back("e");

t.push_back("ffy");

t.push_back("tens");

vector<string> u;

transform(i.begin(), i.end(), j.begin(), back_inserter(k), Plus());

transform(s.begin(), s.end(), t.begin(), back_inserter(u), Plus());

for_each(k.begin(), k.end(), [](int n) { cout << n << " "; });

cout << endl;

for_each(u.begin(), u.end(), [](const string& r) { cout << r << " "; });

cout << endl;

}

C:\Temp>cl /EHsc /nologo /W4 plus.cpp

plus.cpp

C:\Temp>plus

41 52 63

cute fluffy kittens

Compare this to C++98/03 <functional>'s std::plus<T> (which is unchanged in C++0x). Because it's a class template, you'd have to pass plus<int>() and plus<string>(), repeating the element types. Its non-templated function call operator has the form T operator()(const T& x, const T& y) const, making it unable to deal with 2 different types, much less 3 different types, without resorting to implicit conversions. (You can feed plus<string>() a string and a const char *. That will construct a temporary string from the second argument, before concatenating the two strings. The performance of this is not especially desirable.) Finally, because it takes const T&, it can't take advantage of C++0x move semantics. Plus avoids all of this: Plus() doesn't repeat the element type, it deals with the "3 different types" case, and because it uses perfect forwarding, it respects move semantics.

trailing return types

Now, let's look at that templated function call operator again:

template <typename T, typename U>

auto operator()(T&& t, U&& u) const

-> decltype(forward<T>(t) + forward(u)) {

return forward<T>(t) + forward(u);

}

Here, auto has a different meaning from for (auto i = v.begin(); i != v.end(); ++i), where it says "make the type of this thing the same as the type of whatever initializes it". When used as a return type, auto says "this function has a trailing-return-type; after I declare its parameters, I'll tell you what its return type is". (The C++0x Working Draft N2857 calls this a late-specified return type, but this is being renamed to trailing-return-type; see paper N2859.) If this seems suspiciously similar to how lambdas are given explicit return types, that's because it is. A lambda's return type has to go on the right in order for its lambda-introducer [] to appear first. Here, the decltype-powered return type has to go on the right in order for the function parameters t and u to be declared first. Where the auto appears on the left, the template parameters T and U are visible, but the function parameters t and u are not yet visible, and that's what decltype needs. (Technically, decltype(forward<T>(*static_cast<T *>(0)) + forward(*static_cast(0))) could go on the left, but that's an abomination.)

As for the expression given to decltype, giving it the same expression as the return statement ensures correctness in all cases. (Pop quiz: why would decltype(t + u) be wrong?) The repetition here is unavoidable but centralized - it appears exactly once, on adjacent lines, so it is not dangerous.

another example

For completeness, here's that "3 different types" example:

C:\Temp>type mult.cpp

#include <algorithm>

#include <iostream>

#include <iterator>

#include <ostream>

#include <utility>

#include <vector>

using namespace std;

struct Multiplies {

template <typename T, typename U>

auto operator()(T&& t, U&& u) const

-> decltype(forward<T>(t) * forward(u)) {

return forward<T>(t) * forward(u);

}

};

class Watts {

public:

explicit Watts(const int n) : m_n(n) { }

int get() const { return m_n; }

private:

int m_n;

};

class Seconds {

public:

explicit Seconds(const int n) : m_n(n) { }

int get() const { return m_n; }

private:

int m_n;

};

class Joules {

public:

explicit Joules(const int n) : m_n(n) { }

int get() const { return m_n; }

private:

int m_n;

};

Joules operator*(const Watts& w, const Seconds& s) {

return Joules(w.get() * s.get());

}

int main() {

vector<Watts> w;

w.push_back(Watts(2));

w.push_back(Watts(3));

w.push_back(Watts(4));

vector<Seconds> s;

s.push_back(Seconds(5));

s.push_back(Seconds(6));

s.push_back(Seconds(7));

vector<Joules> j;

transform(w.begin(), w.end(), s.begin(), back_inserter(j), Multiplies());

for_each(j.begin(), j.end(), [](const Joules& r) { cout << r.get() << endl; });

}

C:\Temp>cl /EHsc /nologo /W4 mult.cpp

mult.cpp

C:\Temp>mult

You might ask, "is all of this generality really necessary?" The answer is yes, yes it is. I've already mentioned how perfect forwarding and decltype make arithmetic operation functors easier to use (by removing the need to repeat element types), more flexible (by dealing with mixed argument and return types), and more efficient (by respecting move semantics). Essentially, perfect forwarding and decltype allow you to write more "transparent" code. Inflexible code and inefficient code are not transparent - their presence can't be ignored.

advanced rules

decltype is powered by several rules. However, if you stick to the pattern above, they don't matter and it just works. I rarely get to say that about C++, but it's true in this case.

Although the vast majority of decltype uses will follow the pattern above, decltype can be used in other contexts. In that case, you've activated expert mode, and you should read the rules in their entirety. In the C++0x Working Draft N2857, they're given by 7.1.6.2 [dcl.type.simple]/4.

but wait, there's more

decltype is the fifth and final C++0x Core Language feature being added to VC10. While it wasn't in the VC10 CTP, it's in VC10 Beta 1. Also in VC10 Beta 1 are many C++0x Standard Library features, which I'll be blogging about soon!

Stephan T. Lavavej

Visual C++ Libraries Developer

posted on 2011-05-12 00:08 sun_kang 阅读(1220) 评论(0) 编辑收藏举报

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