Question

我很难绕过以下内容（特别是方案b）：（假设我已经定义了赋值运算符，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）

情景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，花花公子

方案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

我的问题是在方案b中，为什么在赋值运算符之前调用复制构造函数是正确的？根据我的理解，只能在未初始化的对象上调用复制构造函数。但是，在这种情况下，对象f已在添加之前的行中初始化。

非常感谢您的解释。

道歉没有立即发布源代码（并且缺少缩进 - 我在复制到textarea时遇到问题）。这里的一切都很简单。我正在使用Visual Studio 2005.不幸的是，我还不熟悉它的工作原理，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的解释是我的新手C ++头脑可以掌握的：）

将优化级别更改为/ O2之后，我可以看到方案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 ，可能会调用复制构造函数。

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;

这是一步一步的分析：

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 `=`

C ++运算符重载和复制构造函数

1 个答案: