C++ 运算符重载和复制构造函数

Question

我很难理解以下内容(特别是场景 b):(假设我已经定义了一个赋值运算符、加法运算符和复制构造函数，只是为了输出它们被调用的事实)

场景一:

Simple a;
Simple b;
Simple c = a + b;

    The output is as follows:
    Simple constructor called
    Simple constructor called
    Simple add operator call
    Simple constructor called
    copy constructor called

-- 这一切都很好，花花公子

场景b(我无法理解的行为):

Simple d;
Simple e;
Simple f;
f = d + e;

    Simple constructor called
    Simple constructor called
    Simple constructor called
    Simple add operator called
    Simple constructor called
    copy constructor called
    assignment operator called

我的问题是，在场景 b 中，为什么在赋值运算符之前调用复制构造函数？据我了解，复制构造函数只会在未初始化的对象上调用。但是，在这种情况下，对象 f 已在添加之前的行中进行了初始化。

如有解释，将不胜感激。

很抱歉没有立即发布源代码(以及缺少缩进 - 我在复制到文本区域时遇到问题)。这就是它的所有简单性。我正在使用 Visual Studio 2005。不幸的是，我还不太熟悉它的工作原理，因此我无法指定传递给编译器的优化参数。

class Simple
{
public:
    Simple(void);
Simple operator +(const Simple& z_Simple) const;
Simple& operator =(const Simple& z_Simple);
Simple(const Simple& z_Copy);
int m_Width;
int m_Height;
public:
~Simple(void);
};


#include "Simple.h"
#include <iostream>

using std::cout;
using std::endl;

Simple::Simple(void)
{
this->m_Height = 0;
this->m_Width = 0;
cout << "Simple constructor called" << endl;
}

Simple::Simple(const Simple& z_Copy)
{
cout << "copy constructor called" << endl;
this->m_Height = z_Copy.m_Height;
this->m_Width = z_Copy.m_Width;
}

Simple& Simple::operator =(const Simple &z_Simple)
{
cout << "assignment operator called" << endl;
this->m_Height = z_Simple.m_Height;
this->m_Width = z_Simple.m_Width;   
return *this;
}


Simple Simple::operator +(const Simple &z_Simple) const
{
cout << "Simple add operator called" << endl;
int y_Height = this->m_Height + z_Simple.m_Height;
int y_Width = this->m_Width + z_Simple.m_Width;
Simple y_Ret;
y_Ret.m_Height = y_Height;
y_Ret.m_Width = y_Width;
return y_Ret;
}

Simple::~Simple(void)
{
cout << "destructor called" << endl;
}

当然，Nemo 的解释是我的 C++ 新手可以理解的:)

将优化级别更改为/O2 后，我可以看到场景 b 的输出如下(以及我所期望的)

    Simple constructor called
    Simple constructor called
    Simple constructor called
    Simple add operator called
    Simple constructor called
    assignment operator called

谢谢大家的建议。

Answer 1

您的 + 运算符按值返回一个对象，如果编译器没有 elide，这可能会导致调用复制构造函数 它。

Simple Simple::operator +(const Simple &z_Simple) const
{
    //......
    Simple y_Ret;
    //......
    return y_Ret;
}

代码:

Simple d;
Simple e;
Simple f;
f = d + e;

下面是一步一步的分析:

Simple constructor called     ---> creation of `d`
Simple constructor called     ---> creation of `e`
Simple constructor called     ---> creation of `f`
Simple add operator called    ---> Inside Addition operator
Simple constructor called     ---> creation of local `y_Ret`
copy constructor called       ---> `y_Ret` returned by value 
assignment operator called    ---> Result returned by `+` used for `=`