c++新特性实验(4)声明与定义:右值引用（C++11）

1.作用

　　c++11以前，临时对象、字面常量一般情况下不可以再次访问，也不可以修改。右值引用可以解决这个问题。

1.1 实验A

#include <iostream>
using namespace std;

class A{
    int     id;
public:
    A(int i) : id(i){
        cout << "A constructor " << id << endl;
    }
    ~A(){
        cout << "A destructor " << id << endl;
    }
    void fun(){
        cout << "I'm " << id << endl;
    }
};

int 
main(int argc,char *argv[]){
    
    char ch = 'a';
    {
        A(1).fun();
        cout << "exit {} " << endl;
    }

    return 0;
}

　　结果：

A constructor 1
I'm 1
A destructor 1
exit {} 

　　问题：

1. 代码第21行和第23行的红色部分，一个是字面常量，一个产生A的临时对象。它们的生命期都是一行。
2. 如果后续代码还想访问它们怎么办？
3. 如果想修改A(1)产生的临时对象的值怎么办？
4. 如果后面还有很多类似A(1).fun()这样的调用，每次都要分配空间，调用构造、析构。如果对象复杂点，那么就很费时。
5. 临时对象生命期结束后，为什么还要去访问、修改它？不频繁的A(1)这样的调用，不就没有问题4了吗？

1.2 C++11以前解决问题2

　　用const引用临时对象或者字面量，修改如下。

int 
main(int argc,char *argv[]){
    
    {
        char ch = 'a';
        A(1).fun();
        cout << "exit {} " << endl;
    }

    {
        const char ch = 'b';
        //A &ref = A(2);    //error,非const左值引用不能引用右值。cannot bind non-const lvalue reference of type ‘A&’ to an rvalue of type ‘A’
        const A &ref = A(2);
        ref.fun();
        cout << "exit {} " << endl;
    }

    return 0;
}

　　同时class A的fun()也要提供const版本。

    void fun() const {
        cout << "I'm " << id << endl;
    }

1.3 c++11以前解决问题3、4

　　无法解决问题3、4

　　c++11起，可以用右值引用解决这两个问题，可以修改临时对象的值，也避免重复分配空间。

 

2.特性（C++11起）

1. 它是右值的引用，右值的别名。
2. 用两个“&&”声明
3. 右值引用只能引用右值（字面常量，临时对象）。
4. 右值引用声明时要初始化,成员右值引用要在构造初始化列表中初始化。
5. 右值引用被列入函数重载规则
6. 它引用的对象在释放前，可以通过它访问、修改(const 右值引用除外)。
7. 它与引用占的空间一样。

2.实验B：特性[1-3]

2.1 左值引用字面量

    char ch = 'a';
    //char &lr = 'e';           //error,非const左值引用不能引用右值
    const char &clr = 'b';      //ok

2.2 右值引用字面量

    char ch = 'b';
    //char && rr2 = ch;                     //error,ch不是右值。
    char && rr1 = 'b';                      //ok
    char && rr2 = 'b' + getInt2();          //ok,普通函数
    char && rr3 = 'b' + 3 * 10 - 28 ^ 1;    //ok
    char && rr4 = 'b' + getInt() / 4;       //ok,constexpr函数
    //char && rr5 = rr4;                    //error,虽然rr4是合法的右值引用，但是它不是右值。

    char *pch = &ch;
    char *&& prch1 = &rr4;                  //ok.
    //char *&& prch2 = pch;                 //error,
    prch1 = &ch;                            //ok.&ch得到ch的地址，它是右值。
    
    cout << " rr1 = " << rr1 << " rr2 = " << rr2 << " rr3 = " << rr3 << " rr4 = " << rr4 << " prch1 = " << *prch1 << endl; 

　　结果

 　　rr1 = b rr2 = l rr3 = e rr4 = i prch1 = b

2.3 右值引用临时对象

    
    A a(0);
    //A &&rr1 = a;                          //error，a不是临时对象
    A(5).fun();
    A && rr2 = {3};
    A &r1 = rr2;                            //ok
    A *pa = &rr2;                           //ok

    rr2.fun();
    r1.fun();
    pa->fun();

    cout << "exit object test {}" << endl;

　　A的定义如下：

class A{
public:
  void fun() const {
        cout << "I'm " << id << endl;
    }
public:
    int     id;
    A(int i) : id(i){
        cout << "A constructor " << id << endl;
    }
    virtual ~A(){
        cout << "A destructor " << id << endl;
    }
    A(const A &a){
        cout << "A copy constructor ,from " << a.id << endl;
        if(&a == this ) return;
        this->id = a.id;
    }
    A& operator=(const A &o){
        cout << "A operator= ,from " << o.id << endl;
        if(&o == this ) return *this;
        this->id = o.id;
        return *this;
    }
    friend ostream& operator<<(ostream &os,const A &a);
};

　　运行结果：

A constructor 0
A constructor 5
I'm 5
A destructor 5
A constructor 3
I'm 3
I'm 3
I'm 3
exit object test {}
A destructor 3
A destructor 0

3.实验C：右值引用的初始化

　　右值引用声明时要初始化,成员右值引用要在构造初始化列表中初始化。

    //声明时要初始化
    char && rr1 ;   //error,未初始化。

　　成员右值引用的初始化。

    class B{
    public:
        char &&rr;
        B():rr('d'){
        }
        B(const B &b):rr('d'){
        }
        B& operator=(const B &b){
            this->rr = b.rr;
            return *this;
        }
        /* error
        char&& getRR(){
            return rr;
        }
        */
    };
    B b ;
    cout << "b.rr = " << b.rr << endl;
    

　　运行结果

　　b.rr = d

4.实验D：右值引用与函数

4.1 右值引用被列入重载参考

class D { };

void 
foo(const D &){
    cout << " foo & called " << endl;
}
void 
foo(const D &&){
    cout << " foo&& called " << endl;
}

　　测试代码

    D d;
    foo(d);
    foo(D());

　　结果：

 　　foo & called 
 　　foo&& called 

　　如果把右值引用重载版本去掉，两个都输出：foo & called

4.2 右值引用与返回值

Rvalue references
Rvalue references can be used to extend the lifetimes of temporary objects (note, 
lvalue references to const can extend the lifetimes of temporary objects too,but they are not modifiable through them):

　　右值引用可以用来为临时对象延长生存期。

　　那它可以延长多久？能超出{}吗？能像堆上的对象的生命期一样？可以把函数返回值声明右值引用然后返回临时对象？

函数结束后，临时对象空间被释放，运行时会出错。

A&&
fun(){
    cout << __func__ << endl;
    return A(9);
}

　　那如果返回全局的右值引用或者成员函数返回成员右值引用?

char && rrrr = 'd';
char &&
fch(){
    return rrrr;
}

　　或者

 class B{
    public:
        char &&rr;
        // error
        char&& getRR(){
            return rr;
        }
        
    };

　　都不可以。

　　只能在它所在的 { } 延长一点生存期、出 } 就释放。

5.实验E：用右值引用修改临时对象

    A && rr2 = {3};
    const A &&rr3 = A(4);                   //ok
    A &r1 = rr2;                            //ok
    A *pa = &rr2;                           //ok

    rr2.fun();
    r1.fun();
    pa->fun();

    rr2.id = 99;

    rr2.fun();
    r1.fun();
    pa->fun();

    char && ch = 'd';
    cout << "ch = " << ch << endl;
    ch = 'e';
    cout << "ch = " << ch << endl;
    
    const char && cch = 'f';
    //cch = 'g';                            //error,const的

6.实验F：右值引用占内存空间吗

6.1 成员引用占空间

在linux 64位、gcc7.4.0 下测试，类里的引用成员与一个指针占空间大小一样。

    class F{
    };
    class E{
    public:
        char && rr;
        E():rr('e'){
        }
    };
    class G{
    public:
        char *pch;
    };
    class H{
    public:
        char &rch;
        H(char ch):rch(ch){
        }
    };
    
    cout << "alignof(F) = " << alignof(F) << "\tsizeof(F) = " << sizeof(F) << endl;
    cout << "alignof(E) = " << alignof(E) << "\tsizeof(E) = " << sizeof(E) << endl;
    cout << "alignof(G) = " << alignof(G) << "\tsizeof(G) = " << sizeof(G) << endl;
    cout << "alignof(H) = " << alignof(H) << "\tsizeof(H) = " << sizeof(H) << endl;

　　结果

alignof(F) = 1    sizeof(F) = 1
alignof(E) = 8    sizeof(E) = 8
alignof(G) = 8    sizeof(G) = 8
alignof(H) = 8    sizeof(H) = 8

　　win10_64 + vs2019 结果：

alignof(F) = 1  sizeof(F) = 1
alignof(E) = 4  sizeof(E) = 4
alignof(G) = 4  sizeof(G) = 4
alignof(H) = 4  sizeof(H) = 4

6.2 验证G:引用是指针？

　　如果引用是const指针，那么它应该有独立的地址。

    int a = 10;
    int &ra = a;
    const int * const pa = &a;
    
    cout << " a.addr = " << &a  << endl;
    cout << "ra.addr = " << &ra << endl;
    cout << "pa.addr = " << &pa << endl;
    cout <<"(*pa).addr = " << pa << endl;

　　结果

　　 a.addr = 0x7fff3810df64
　　ra.addr = 0x7fff3810df64
　　pa.addr = 0x7fff3810df68
　　(*pa).addr = 0x7fff3810df64

　　如果ra是个类似pa一样的const指针，它应该和pa一样，有自己的地址，但是对ra取地址时，它和a的地址一样。普通引用不占空间。　

7.完整示例

  

#include <iostream>
using namespace std;

class A{
public:
  void fun() const {
        cout << "I'm " << id << endl;
    }
public:
    int     id;
    A(int i) : id(i){
        cout << "A constructor " << id << endl;
    }
    virtual ~A(){
        cout << "A destructor " << id << endl;
    }
    A(const A &a){
        cout << "A copy constructor ,from " << a.id << endl;
        if(&a == this ) return;
        this->id = a.id;
    }
    A& operator=(const A &o){
        cout << "A operator= ,from " << o.id << endl;
        if(&o == this ) return *this;
        this->id = o.id;
        return *this;
    }
    friend ostream& operator<<(ostream &os,const A &a);
};
ostream& operator<<(ostream &os,const A &a){
    os << "A.id = " << a.id << endl;
    return os;
}

constexpr int
getInt(){
    return 30;
}
int getInt2(){
    int a = 10;
    return a;
}

class D { };

void 
foo(const D &){
    cout << " foo & called " << endl;
}
void 
foo(const D &&){
    cout << " foo&& called " << endl;
}


A&&
fun(){
    cout << __func__ << endl;
    return A(9);
}

/*
char && rrrr = 'd';
char &&
fch(){
    return rrrr;
}
*/
int 
main(int argc,char *argv[]){
    
    {
        char ch = 'a';
        //char &lr = 'e';                       //error,非const左值引用不能引用右值
        const char &clr = 'b';                  //ok
    }
    {
        char ch = 'b';
        //char && rr2 = ch;                     //error,ch不是右值。
        char && rr1 = 'b';                      //ok
        char && rr2 = 'b' + getInt2();          //ok,普通函数
        char && rr3 = 'b' + 3 * 10 - 28 ^ 1;    //ok
        char && rr4 = 'b' + getInt() / 4;       //ok,constexpr函数
        //char && rr5 = rr4;                    //error,虽然rr4是合法的右值引用，但是它不是右值。

        char *pch = &ch;
        char *&& prch1 = &rr4;                  //ok.
        //char *&& prch2 = pch;                 //error,
        prch1 = &ch;                            //ok.&ch得到ch的地址，它是右值。
        
        cout << " rr1 = " << rr1 << " rr2 = " << rr2 << " rr3 = " << rr3 << " rr4 = " << rr4 << " prch1 = " << *prch1 << endl; 
    }
    {
        A a(0);
        //A &&rr1 = a;                          //error，a不是临时对象
        A(5).fun();
        A && rr2 = {3};                         //ok
        const A &&rr3 = A(4);                   //ok
        A &r1 = rr2;                            //ok
        A *pa = &rr2;                           //ok

        rr2.fun();
        r1.fun();
        pa->fun();

        cout << "exit object test {}" << endl;
    }
    {
        //声明时要初始化
        //char && rr1 ;                         //error,未初始化。
        
        class B{
        public:
            char &&rr;
            /* error
            char&& getRR(){
                return rr;
            }
            */
            B():rr('d'){
            }
            B(const B &b):rr('d'){
            }
            B& operator=(const B &b){
                this->rr = b.rr;
                return *this;
            }
           
            /* error
            char&& getRR2(){
                return rrrr;
            }
            */
        };
        B b ;
        cout << "b.rr = " << b.rr << endl;
        
    }
    {
        D d;
        foo(d);
        foo(D());


        A && rr = fun();
        //cout << " rr = " << rr << endl;
    }

    {
        A && rr2 = {3};
        const A &&rr3 = A(4);                   //ok
        A &r1 = rr2;                            //ok
        A *pa = &rr2;                           //ok

        rr2.fun();
        r1.fun();
        pa->fun();

        rr2.id = 99;

        rr2.fun();
        r1.fun();
        pa->fun();

        pa->id = 3;

        rr2.fun();
        r1.fun();
        pa->fun();


        char && ch = 'd';
        cout << "ch = " << ch << endl;
        ch = 'e';
        cout << "ch = " << ch << endl;
        
        const char && cch = 'f';
        //cch = 'g';                            //error,const的
    }


    {
        class F{
        };
        class E{
        public:
            char && rr;
            E():rr('e'){
            }
        };
        class G{
        public:
            char *pch;
        };
        class H{
        public:
            char &rch;
            H(char ch):rch(ch){
            }
        };
        
        cout << "alignof(F) = " << alignof(F) << "\tsizeof(F) = " << sizeof(F) << endl;
        cout << "alignof(E) = " << alignof(E) << "\tsizeof(E) = " << sizeof(E) << endl;
        cout << "alignof(G) = " << alignof(G) << "\tsizeof(G) = " << sizeof(G) << endl;
        cout << "alignof(H) = " << alignof(H) << "\tsizeof(H) = " << sizeof(H) << endl;
    }

    {
        int a = 10;
        int &ra = a;
        const int * const pa = &a;
        
        cout << " a.addr = " << &a  << endl;
        cout << "ra.addr = " << &ra << endl;
        cout << "pa.addr = " << &pa << endl;
        cout <<"(*pa).addr = " << pa << endl;

    }
    return 0;
}