C ++复制构造函数行为 [英] C++ copy constructor behaviour

问题描述

有一部分C ++代码我真的不明白。
我也不知道在哪里搜索关于它的信息，所以我决定提出一个问题。

There is a part of C++ code I don't really understand. Also I don't know where should I go to search information about it, so I decided to ask a question.

#include <iostream>
#include <string>

using namespace std;

class Test
{
    public:
        Test();
        Test(Test const & src);
        Test& operator=(const Test& rhs);
        Test test();
        int x;
};

Test::Test()
{
    cout << "Constructor has been called" << endl;
}

Test::Test(Test const & src)
{
    cout << "Copy constructor has been called" << endl;
}

Test& Test::operator=(const Test& rhs)
{
    cout << "Assignment operator" << endl;
}

Test Test::test()
{
    return Test();
}

int main()
{
    Test a;
    Test b = a.test();

    return 0;
}

为什么我得到的输入是

Constructor has been called
Constructor has been called

？
a.test（）通过调用Test（）创建一个新实例，这就是显示第二条消息的原因。但为什么没有复制构造函数或赋值调用？
如果我将return Test（）更改为return *（new Test（）），那么将调用复制构造函数。

? a.test() creates a new instance by calling "Test()" so that's why the second message is displayed. But why no copy constructor or assignment called? And if I change "return Test()" to "return *(new Test())" then the copy constructor is called.

它是第一次吗？

推荐答案

编译器非常聪明。两个副本 - 从 test 返回并初始化 b （不是此分配）当没有被绑定到引用（12.2）的临时类对象将被引用时，规则（C ++ 11§12.8）：

Compilers are very smart. Both copies - returning from test and initialising b (not this is not an assignment) - are elided according to the following rule (C++11 §12.8):

复制/移动到具有相同cv非限定类型的类对象，可以通过将临时对象直接构造到省略的复制/移动的目标中来省略复制/移动操作

when a temporary class object that has not been bound to a reference (12.2) would be copied/moved to a class object with the same cv-unqualified type, the copy/move operation can be omitted by constructing the temporary object directly into the target of the omitted copy/move

编译器允许这样做，即使它会改变程序的行为（如删除您的输出消息）。预期您不会编写具有其他副作用的复制/移动构造函数和赋值运算符。

Compilers are allowed to do this even if it would change the behaviour of your program (like removing your output messages). It's expected that you do not write copy/move constructors and assignment operators that have other side effects.

请注意，这只是可能发生复制删除的四种情况之一不计算as-if规则）。

Note that is only one of four cases in which copy elision can occur (not counting the as-if rule).

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