C ++运算符重载：没有从对象到引用的已知转换？ [英] C++ operator overloading: no known conversion from object to reference?

问题描述

当我尝试编译以下（g ++ 4.6.3）

When I try to compile the following (g++ 4.6.3)

class A {};

A& operator*=( A& a, const A& b )
{
  return a;
}

A operator*( const A& a, const A& b )
{
  return A( a ) *= b;
}

int main( int, char*[] )
{
  A a, b;

  a = a*b;

  return 0;
}

我得到错误

/tmp/test.cxx: In function ‘A operator*(const A&, const A&)’:
/tmp/test.cxx:14:20: error: no match for ‘operator*=’ in ‘(* & a) *= b’
/tmp/test.cxx:14:20: note: candidate is:
/tmp/test.cxx:6:1: note: A& operator*=(A&, const A&)
/tmp/test.cxx:6:1: note:   no known conversion for argument 1 from ‘A’ to ‘A&’

这让我很困惑 - 如何从一个类转换为该类的引用不知道？

This puzzles me - how can a conversion from a class to a reference to that class not be known?

更改类A的声明如下所示：

Changing the declaration of class A as follows does not have any effect:

class A
{
public:
  A() {}
  A( const A& ) {}
};

同样的错误。

推荐答案

像Lucian说的，你不能将一个临时对象绑定到一个非临时对象上， const引用。编译器的期望是，对象将在表达式之后停止存在，所以修改它是没有意义的。

Like Lucian said, you cannot bind a temporary object to a non-const reference. The expectance of the compiler is that the object will cease to exist after the expression so it makes no sense to modify it.

要修复你的代码，删除临时在 operator * = ）中的参数 const& 没有意义：

To fix your code, remove the temporary (making the argument const& makes no sense in operator *=):

A operator*(A a, const A& b)
{
    return a *= b;
}

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