C ++ operator =返回引用* this [英] C++ operator= return reference to *this

问题描述

查看以下代码：

  #include< iostream> 
 using namespace std; 
 
 class Widet {
 public：
 Widet（int val = 0）：value（val）
 {
 
} $ b b 
 Widet& operator =（Widet& rhs）
 {
 value = rhs.value; 
 return * this; 
} 
 int getValue（）
 {
返回值; 
} 
 private：
 int value; 
}; 
 
 int main（）
 {
 Widet obj1（1）; 
 Widet obj2（2）; 
 Widet obj3（0）; 
（obj3 = obj2）= obj1; 
 cout<< obj3 =< obj3.getValue（）<< endl 
} 
  

代码成功运行，输出为（使用VS2008）：

>

当我让运算符=返回一个值而不是引用：

  Widet operator = ; rhs）
 {
 value = rhs.value; 
 return * this; 
} 
  

它也会成功运行，输出是：

我的问题是：为什么第二个代码运行良好？我们不应该得到一个错误？

为什么是一个好习惯返回引用* * this？

解决方案

为什么第二个代码运行良好？ / p>

因为它是完全有效的代码。它返回对象的临时副本，并且允许在临时对象上调用成员函数（包括 operator =（）），因此没有错误。 p>

如果对象不可复制，您会收到错误。

为什么是好习惯返回引用* this而不是* this？

因为不是所有的对象都是可复制的， 。你可以引用任何对象，引用总是很便宜的。

Look at the following code:

#include <iostream>
using namespace std;

class Widet{
public:
    Widet(int val = 0):value(val)
    {

    }

    Widet& operator=(Widet &rhs)
    {
        value = rhs.value;
        return *this;
    }
    int getValue()
    {
        return value;
    }
private:
    int value;
};

int main()
{
    Widet obj1(1);
    Widet obj2(2);
    Widet obj3(0);
    (obj3 = obj2) = obj1;
    cout << "obj3 = " << obj3.getValue() << endl;
}

The code runs successfully and the output is (using VS2008):

When I let the operator= return a value instead of reference:

Widet operator=(Widet &rhs)
{
    value = rhs.value;
    return *this;
}

It also runs successfully and the output is :

My question is :Why the second code runs well?Should not we get a error?

Why it is a good habit to return reference to *this instead of *this?

解决方案

Why the second code runs well?Should not we get a error?

Because it's perfectly valid code. It returns a temporary copy of the object, and you're allowed to call member functions (including operator=()) on temporary objects, so there is no error.

You would get an error if the object were uncopyable.

Why it is a good habit to return reference to *this instead of *this?

Because not all objects are copyable, and some objects are expensive to copy. You can take a reference to any object, and references are always cheap to pass around.

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