Ambiguous operator overload?

[tavianator]

I have an old C++ project which I recently tried use after not compiling it for a while, only to find that g++ borked on it. The problem is reducible to this:

class A
{
public:
  A() { }
  template <typename T> A(T t) { }
};
 
template <typename T>
A
operator-(const A& lhs, T rhs)
{ return A(); }
 
class B
{
public:
  B() { }
 
  B operator-(unsigned int rhs) { return B(); }
};
 
int
main()
{
  B b;
  b - 1;
  return 0;
}
 

g++ now complains that "b - 1" is ambiguous, because the C++ standard says so even though the worst conversion for B::operator-(unsigned int) is better than the worst conversion for operator-(const A&, T). Is this true?

[Banfa]

It's true and it is also what you would want. The compiler is acting unintelligently, it is saying neither operation is an exact match therefore I will not compile this, you will have to explicitly state which conversion you want to happen.

There are a number of ways to do this, in your simple example change 1 to 1U, this means no conversion is required for the B::operator- so it is an exact match and is used.

Or you could explicitly invoke the operator by changing b - 1 to

b.operator-(1);

However the point is that the compiler does secretly make any decisions for you when there is an ambiguity (which is personally how I like all my software not just compilers).

[weaknessforcats]

Your class A needs to be a template.

template<class T>  
class A 
{ 
public: 
  A() { } 
  template <typename T> A(T t) { } 
 
}; 
 template<class T> 
A<T> operator-(const A<T>& lhs, T rhs) 
{ return A(); } 
 

As initially written the compiler can't tell whether to use the inline function of class B:

B operator-(unsigned int rhs) { return B(); } 
 

or to use

template <typename T> 
A 
operator-(const A& lhs, T rhs) 
{ return A(); } 
 

where T is a object of class B or whether to use

By making class A a template, the ambiguity is resolved since

B operator-(unsigned int rhs) { return B(); } 
 

is a better conversion than

 template<class T> 
A<T> operator-(const A<T>& lhs, T rhs) 
{ return A(); } 
 

where T is an object class B.

[tavianator]

@weaknessforcats
But in this case, all A's need to have the same type, no matter what type T was used in its template constructor, or what possibly different type T is subtracted from it.

@weaknessforcats
Perhaps you meant:

 template<class T, class U> 
A<T> operator-(const A<T>& lhs, U rhs) 
{ return A<T>(); } 
 

Either way, I can't really do that. But why is it that the original code is ambiguous? I always thought that overloads were ranked by the worst conversion necessary, and since the worst conversion for B::operator-(unsigned int) (int -> unsigned int) is better than the worst conversion for operator-(const A&, T), T = int (B -> A via user-defined constructor), where is the ambiguity?

[weaknessforcats]

When your class A is not a template, there is still a function template for the A constructor. Therefore, when you b-1 the compiler could call:

template <typename T> 
A 
operator-(const A& lhs, T rhs) 
{ return A(); } 
 

The lhs argument could obtained as an A object intialized by a B object using the template A constructor.

Or the compiler could call:

class B 
{ 
public: 
  B() { } 
 
  B operator-(unsigned int rhs) { return B(); } 
}; 
 

Hence the ambiguity.

When class A is a template then class B is preferred to A<T>.