C++ operator overloading: no known conversion from object to reference?

问题

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;
}

I get the error

/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&’

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

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

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

Same error.

I would be extremely grateful for hints as to what's going on here.

回答1:

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.

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;
}

回答2:

When you write A( a ), you create a temporary of type A ( a rvalue ) that you copy-construct with a. C++ states that no rvalue could be passed as non const reference. Visual Studio is a bit sloppy about this rule, but gcc and the like enforce it.

To fix, try this ( which is exactly the same, but you create a lvalue by naming that variable ). More on l- and r-value here

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

来源：https://stackoverflow.com/questions/10236240/c-operator-overloading-no-known-conversion-from-object-to-reference

标签

c++

operator-overloading