对于两个或多于两个对象的C ++中的重载赋值运算符？ [英] Overloaded Addition assignment operator in C++ for two /more than two objects?

问题描述

我已将此运算符重载。

class sample
{
private : 
  int x;
public :
  sample(int x1 =0)
  {
    x = x1;
  }

  sample operator+(sample s);
};

sample sample::operator+(sample s)
{
  x = x + s.x;
  return *this;
}

int  main()
{
  sample s1(10);
  sample s2;
  s2 = s2 + s1;
  return 0;    
}

这是否正确？
我的问题是如果我想添加两个不同的示例对象如何重载opeartor; s = s1 + s2 ;

推荐答案

使用friend运算符重载应该为你的技巧并且是定义二元运算符的常用方法，只需添加：

Using friend operator overload should do the trick for you and is a common way to define binary operators, just add:

friend sample operator+(const sample& a, const sample& b); //in class

sample operator+(const sample& a, const sample& b) { //outside the class
    return sample(a.x + b.x);
}

如果您希望它保留为会员，方案，没有颠倒），你必须让运算符 const 函数：

If you want it to remain a member, (which has downsides in some rare scenarios, and no upsides), you have to make the operator a const function:

sample operator+(sample s) const; //in class

sample sample::operator+(const sample& b) const { //outside the class
    return sample(this->x + b.x);
}

其中任何一个都允许操作符链接。您之前的 s = s + s1 + s2 失败的原因是 s + s1 会执行并返回一个临时 示例对象。然后，它将尝试向该示例添加 s2 。但是暂时只能使用 const 引用 ^[1] ，因此只能使用 const 成员函数。由于运算符+ 成员函数不是 const 函数，因此不能在 const 临时。注意，为了使它 const ，我不得不重写它，因为你的版本修改了 + 。

Either of these will allow operator chaining. The reason your previous s = s + s1 + s2 was failing, is that the s + s1 would execute and return a temporary sample object. Then, it would attempt to add s2 to that sample. However, temporaries can only be const references^[1], and as such, can only use const member functions. Since your operator+ member function is not a const function, you cannot use that function on the const temporary. Note that to make it const, I had to rewrite it, since your version modifies the object on the left side of the +.

^{[1]与异常不是特别相关，即rvalues}

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