表达式MyClass a = b + c是一个拷贝初始化吗？

Question

我知道表达式像MyClass a = b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，似乎表达式不是复制初始化，也不是复制任务。所以它是什么？

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; 
} 

并且输出是

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

Answer 1

你体验被称为返回值优化什么（见What are copy elision and return value optimization?）：

这是在这里的第一个例子描述。

编译器优化代码（以最小化创建的对象），并使得a实际上是您的+运算符声明并返回的本地对象。然后，没有使用复制构造函数。这是你在这里观察到的。

很好地观察到这一点，改变你的代码，看看对象的内存不会忽略：

#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 

你看，在operator+创建的临时对象是一样的a（0x7ffd44b769ff ），由于编译器的优化。

如果使用g ++，请使用-fno-elide-constructors进行编译以禁用此编译器优化。而现在输出的是：

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，现在operator+创建一个本地对象（0x7ffd92847cff），这是后来复制的临时对象return语句（复制0x7ffd92847cff到0x7ffd92847d1f），最后将其用于构建通过复制（复制0x7ffd92847d1f至0x7ffd92847d1e）。

Answer 2

在这样的情况下的复制构造may be elided。

Answer 3

从C++型perslective，有因为operator+返回一个rvalue MyClass的

MyClass a = b + c; 

和

MyClass a = MyClass(); 

之间没有差别。而且由于您刚刚创建了您想要的对象a，因此它只是将该对象直接放入a。 http://en.cppreference.com/w/cpp/language/copy_elision