C++运算符重载和复制构造函数

C++ operator overloading and the copy constructor

本文关键字：复制 构造函数 重载 运算符 C++      更新时间：2023-10-16

我很难理解以下内容（特别是场景b）：（假设我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a:

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，很好的

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，为什么复制构造函数在赋值运算符之前被调用？据我所知，复制构造函数只会在未初始化的对象上调用。然而，在这种情况下，对象f已经在添加之前的行中初始化。

如有解释，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）。它就在这里，一切都很简单。我使用的是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，尼莫的解释是我的C++新手能够理解的：）

在将优化级别更改为/O2之后，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。

您的+运算符按值返回一个对象，如果编译器没有消除它，可能会导致对复制构造函数的调用。

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;

以下是一个逐步分析：

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 `=`
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