【C++编程】std::forward与完美转发
std::forward 完美转发
1. std::forward代码原型:
template <class T>
T&& forward(typename remove_reference<T>::type& t) noexcept {
return static_cast<T&&>(t);
}
template <class T>
T&& forward(typename remove_reference<T>::type&& t) noexcept {
return static_cast<T&&>(t);
}
2. 分析std::forward<T>实现条件转发的原理(以转发Widget类对象为例)
当传递给func
函数的实参类型为左值Widget
时,T被推导为Widget &
,然后forward会实例化为std::forward<Widget &>
,并返回Widget&
(左值引用,根据定义是个左值!)而当传递给func
函数的实参类型为右值Widget
时,T被推导为Widget
。然后forward被实例化为std::forward<Widget>
,并返回Widget&&
(注意,匿名的右值引用是个右值!)
可见,std::forward
会根据传递给func
函数实参(注意,不是形参)的左/右值类型进行转发。当传给func函数左值实参时,forward返回左值引用,并将该左值转发给process。而当传入func的实参为右值时,forward返回右值引用,并将该右值转发给process函数。
1 #include <iostream>
2 using namespace std;
3
4 template <typename Tp>
5 inline typename std::remove_reference<_Tp>::type &&my_move(Tp &&t)
6 {
7 cout << &t << endl;
8 return static_cast<typename std::remove_reference<Tp>::type &&>(t);
9 }
10
11 template <typename Tp>
12 inline Tp &&my_forward(typename std::remove_reference<Tp>::type &t)
13 {
14 cout << "calling lvalue forward" << endl;
15 return static_cast<Tp &&>(t);
16 }
17
18 template <typename Tp>
19 inline Tp &&my_forward(typename std::remove_reference<Tp>::type &&t)
20 {
21 cout << "calling rvalue forward" << endl;
22 static_assert(!std::is_lvalue_reference<_Tp>::value,
23 "template argument substituting Tp is an lvalue reference type");
24 return static_cast<Tp &&>(t);
25 }
26
27 struct X
28 {
29 };
30
31 void inner(const X &) { cout << "inner(const X&)" << endl; }
32 void inner(X &&) { cout << "inner(X&&)" << endl; }
33 template <typename T>
34 void outer(T &&t)
35 {
36 inner(my_forward<T>(t));
37 }
38
39 int main()
40 {
41 cout << ">>> test move" << endl;
42 X a1;
43 cout << &a1 << endl;
44 cout << "> test lvalue" << endl;
45 X &&b = my_move(a1);
46 cout << "> test rvalue" << endl;
47 X &&c = my_move(X());
48
49 cout << endl;
50
51 cout << ">>> test forward" << endl;
52 X a2;
53 cout << "> test lvalue indirectly" << endl;
54 outer(a2);
55 cout << "> test rvalue indirectly" << endl;
56 outer(X());
57 cout << "> test rvalue directly" << endl;
58 inner(my_forward<X>(X()));
59
须知,std::move
和 std::forward
都是仅仅执行强制型别转换的函数,std::move
无条件的将实参强制转换成右值, 而std::forward
则 仅在某个特定的条件满足时才执行同一类型转换。
示例:
#include <iostream>
#include <memory>
#include <utility>
struct A {
A(int&& n) { std::cout << "rvalue overload, n=" << n << '\n'; }
A(int& n) { std::cout << "lvalue overload, n=" << n << '\n'; }
};
class B {
public:
template<class T1, class T2, class T3>
B(T1&& t1, T2&& t2, T3&& t3)
:a1_{std::forward<T1>(t1)}, a2_{std::forward<T2>(t2)}, a3_{std::forward<T3>(t3)} {}
private:
A a1_, a2_, a3_;
};
template<class T, class U>
std::unique_ptr<T> make_unique1(U&& u) {
return std::unique_ptr<T>(new T(std::forward<U>(u)));
}
template<class T, class... U>
std::unique_ptr<T> make_unique2(U&&... u) {
return std::unique_ptr<T>(new T(std::forward<U>(u)...));
}
auto make_B(auto&&... args) { //since C++20
return B(std::forward<decltype(args)>(args)...);
}
int main()
{
auto p1 = make_unique1<A>(2); // rvalue
int i = 1;
auto p2 = make_unique1<A>(i); // lvalue
std::cout << "B\n";
auto t = make_unique2<B>(2, i, 3);
std::cout << "make_B\n";
[[maybe_unused]] B b = make_B(4, i, 5);
}
输出:
rvalue overload, n=2
lvalue overload, n=1
B
rvalue overload, n=2
lvalue overload, n=1
rvalue overload, n=3
make_B
rvalue overload, n=4
lvalue overload, n=1
rvalue overload, n=5