复制构造函数不被调用？ [英] copy constructor is not called?

问题描述

class X
{
  int i;  
  public:
  X(int m) : i(m) {};

  X(const X& x)
  {
    //cout "copy constructor is called\n";
  }

  const X opearator++(X& a,int)
  {
     //cout "X++ is called\n";
     X b(a.i);
     a.i++;
     return b;
  }
  void f(X a)
  {   }
};

 int main()
{ 
  X a(1);
  f(a);
  a++; 
  return 0;
}

这里当调用函数'f'时，拷贝构造函数会按预期调用。在++的情况下，调用operator ++函数，但是当它返回copy constructor is not called时。
为什么从函数operator ++返回时不会调用copy contructor？

Here when function 'f' is called copy constructor is getting called as expected. In case of a++, operator++ function is called but when it returns "copy constructor is not called". why "copy contructor is not called while returning from function 'operator++'?

推荐答案

返回值优化（RVO）
http://en.wikipedia.org/wiki/Return_value_optimization

I believe you've encountered return value optimization (RVO) http://en.wikipedia.org/wiki/Return_value_optimization

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