c++ - Implicit conversion when overloading operators for template classes

Question

I would like to know why implicit type conversion doesn't work with outside operator overloading on class templates. Here is the working, non-templated version:

class foo
{
public:

    foo() = default;

    foo(int that)
    {}

    foo& operator +=(foo rhs)
    {
        return *this;
    }
};

foo operator +(foo lhs, foo rhs)
{
    lhs += rhs;
    return lhs;
}

As expected, the following lines compile correctly:

foo f, g;
f = f + g; // OK
f += 5; // OK
f = f + 5; // OK
f = 5 + f; // OK

On the other hand, when class foo is declared as a simple template like this:

template< typename T >
class foo
{
public:

    foo() = default;

    foo(int that)
    {}

    foo& operator +=(foo rhs)
    {
        return *this;
    }
};

template< typename T >
foo< T > operator +(foo< T > lhs, foo< T > rhs)
{
    lhs += rhs;
    return lhs;
}

The following lines compile with errors:

foo< int > f, g;
f = f + g; // OK
f += 5; // OK
f = f + 5; // Error (no match for operator+)
f = 5 + f; // Error (no match for operator+)

I would like to understand why the compiler (GCC 4.6.2) is unable to perform implicit type conversion using the converting constructor for the template version of the class. Is that the expected behaviour? Apart from manually creating all the necessary overloads, is there any workaround for this?

Answer 1

The reason it does not just work is that implicit type conversions (that is, via constructors) do not apply during template argument deduction. But it works if you make the outside operator a friend since then the type T is know, allowing the compiler to investigate what can be casted to make the arguments match.

I made an example based on yours (but removed C++11 stuff), inspired by Item 46 (a rational number class) in Scott Meyers Effective C++ (ed 3). Your question is almost an exact match to that item. Scott also notes that ... "this use of friend is not related to the access of non-public parts of the class."

This will also allow work with mixes of foo< T >, foo< U > etc as long as T and U can be added etc.

Also look at this post: C++ addition overload ambiguity

#include <iostream>

using namespace std;

template< class T >
class foo
{
private:
   T _value;
public:
   foo() : _value() {}

   template <class U>
   foo(const foo<U>& that) : _value(that.getval()) {}

   // I'm sure this it can be done without this being public also;
   T getval() const { return _value ; }; 

   foo(const T& that) : _value(that) {}

   friend const foo operator +(foo &lhs,const foo &rhs) 
      {
     foo result(lhs._value+rhs._value); 
     return result;
      };
   friend const foo operator +(foo &lhs,const T &rhsval) 
      {
     foo result(lhs._value+rhsval); 
     return result;
      };
   friend const foo operator +(const T &lhsval,foo &rhs) 
      {
     foo result(lhsval+rhs._value); 
     return result;
      };

   friend foo& operator +=(foo &lhs,const foo &rhs)
      {
     lhs._value+=rhs._value;

     return lhs;
      };   
   friend std::ostream& operator<<(std::ostream& out, const foo& me){
      return out <<me._value;
   }
};

int main(){
   foo< int > f, g;
   foo< double > dd;
   cout <<f<<endl;
   f = f + g;
   cout <<f<<endl;
   f += 3 ;
   cout <<f<<endl;
   f = f + 5;
   cout <<f<<endl;
   f = 7 + f; 
   cout <<f<<endl;      
   dd=dd+f;
   cout <<dd<<endl;      
   dd=f+dd;
   cout <<dd<<endl;      
   dd=dd+7.3;
   cout <<dd<<endl;             
}