表達式MyClass a = b + c是一個拷貝初始化嗎？

Question

我知道表達式像MyClass a = b是一個複製初始化，它會調用複製構造函數。

MyClass a = b + c呢？

我試過了。在Visual Studio 2015中，似乎表達式不是複製初始化，也不是複製任務。所以它是什麼？

下面是我測試的代碼：

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; 
} 

並且輸出是

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

Answer 1

你體驗被稱爲返回值優化什麼（見What are copy elision and return value optimization?）：

這是在這裏的第一個例子描述。

編譯器優化代碼（以最小化創建的對象），並使得a實際上是您的+運算符聲明並返回的本地對象。然後，沒有使用複製構造函數。這是你在這裏觀察到的。

很好地觀察到這一點，改變你的代碼，看看對象的內存不會忽略：

#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; 
} 

它輸出：

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+創建的臨時對象是一樣的a（0x7ffd44b769ff ），由於編譯器的優化。

如果使用g ++，請使用-fno-elide-constructors進行編譯以禁用此編譯器優化。而現在輸出的是：

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，現在operator+創建一個本地對象（0x7ffd92847cff），這是後來複製的臨時對象return語句（複製0x7ffd92847cff到0x7ffd92847d1f），最後將其用於構建通過複製（複製0x7ffd92847d1f至0x7ffd92847d1e）。

Answer 2

在這樣的情況下的複製構造may be elided。

Answer 3

從C++型perslective，有因爲operator+返回一個rvalue MyClass的

MyClass a = b + c; 

和

MyClass a = MyClass(); 

之間沒有差別。而且由於您剛剛創建了您想要的對象a，因此它只是將該對象直接放入a。 http://en.cppreference.com/w/cpp/language/copy_elision