CODEWITHSUNDEEP
c++operator-overloading
DrGerm
DrGerm

Reputation: 275

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

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.

Upvotes: 13

Views: 15338

Answers (2)

crazyjul
crazyjul

Reputation: 2539

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;
}

Upvotes: 5

Konrad Rudolph
Konrad Rudolph

Reputation: 545508

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;
}

Upvotes: 13

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