表达式 MyClass a = b + c 是复制初始化吗？

Question

我知道像 MyClass a = b 这样的表达式是一个复制初始化，它会调用复制构造函数。

MyClass a = b + c 呢？

我试过了。在visual studio 2015 中，表达式似乎不是复制初始化也不是复制assignment.So 是什么？

下面是我测试的代码：

class MyClass
{
public:
    MyClass()
    {
        cout << "default constructor has been called!" << endl;
    }
    MyClass(const MyClass& cls)
    {
        cout << "copy constructor has been called!" << endl;
    }
    MyClass operator+(const MyClass& cls) const
    {
        cout << "operator + has been called!" << endl;
        return MyClass();
    }
    MyClass& operator=(const MyClass& cls)
    {
        cout << "operator = has been called!" << endl;
        return *this;
    }
};

int main()
{
    MyClass b, c;
    MyClass a = b + c;
    return 0;
}

输出为

default constructor has been called!
default constructor has been called! 
operator + has been called!
default constructor has been called!

Answer 1

这种情况下的复制构造函数may be elided。

Answer 2

从c++类型的角度来看，

没有区别

MyClass a = b + c;

和

MyClass a = MyClass();

因为 operator+ returns 右值 MyClass。由于您刚刚创建了您希望 a 成为的对象，它只是将该对象直接变成 a。

这在此处的第一个示例中进行了描述：http://en.cppreference.com/w/cpp/language/copy_elision

Answer 3

你所体验的叫做return价值优化（参见What are copy elision and return value optimization?）。

编译器优化代码（以最小化创建的对象）并使 a 实际上是声明的本地对象，并由您的 + 运算符 return 编辑。然后，没有使用复制构造函数。这就是您在这里观察到的。

为了很好地观察这一点，更改代码以查看对象内存地址：

#include <iostream>
using namespace std;

class MyClass
{
public:
    MyClass()
    {
        cout << "default constructor has been called for " << std::hex << this << endl;
    }
    MyClass(const MyClass& cls)
    {
        cout << "copy constructor has been called for " << std::hex << this << " (copied from " << &cls << ")" << endl;
    }
    MyClass operator+(const MyClass& cls) const
    {
        cout << "operator + has been called!" << endl;
        MyClass res;
        cout << "operator + returns temporary object " << std::hex << &res << endl;
        return res;
    }
    MyClass& operator=(const MyClass& cls)
    {
        cout << "operator = has been called!" << endl;
        return *this;
    }
};

int main()
{
    MyClass b, c;
    MyClass a = b + c;
    cout << "a object is " << std::hex << &a << endl;
    return 0;
}

它输出：

default constructor has been called for 0x7ffd44b769fd
default constructor has been called for 0x7ffd44b769fe
operator + has been called!
default constructor has been called for 0x7ffd44b769ff
operator + returns temporary object 0x7ffd44b769ff
a object is 0x7ffd44b769ff

由于编译器优化，您看到在 operator+ 中创建的临时对象与 a (0x7ffd44b769ff) 相同。

如果您使用 g++，请使用 -fno-elide-constructors 进行编译以禁用此编译器优化。现在的输出是：

default constructor has been called for 0x7ffd92847d1c
default constructor has been called for 0x7ffd92847d1d
operator + has been called!
default constructor has been called for 0x7ffd92847cff
operator + returns temporary object 0x7ffd92847cff
copy constructor has been called for 0x7ffd92847d1f (copied from 0x7ffd92847cff)
copy constructor has been called for 0x7ffd92847d1e (copied from 0x7ffd92847d1f)
a object is 0x7ffd92847d1e

你看到现在 operator+ 创建了一个本地对象 (0x7ffd92847cff)，稍后将其复制为 return 语句的临时对象（将 0x7ffd92847cff 复制到 0x7ffd92847d1f), 最后用来构造一个by copy(copying 0x7ffd92847d1f to 0x7ffd92847d1e).

表达式 MyClass a = b + c 是复制初始化吗？

Is the expression MyClass a = b + c a copy initialization?

c++

grammar

visual-studio