Deduce return type of operator/function for templates

Question

Is something like this possible?

// We can even assume T and U are native C++ types
template<typename T, typename U>
magically_deduce_return_type_of(T * U) my_mul() { return T * U; }

Or would somebody have to hack up a return_type struct and specialize it for every pair of native types?

Answer 1

Heard of decltype?

In C++0x you can do

template<class T, class U>
auto mul(T x, U y) -> decltype(x*y)
{
    return x*y;
}

Answer 2

You can do this in non C++0x code:

template<typename T, typename U> class Mul
{
  T t_;
  U u_;
public:
  Mul(const T& t, const U& u): t_(t), u_(u) {}
  template <class R>
  operator R ()
  {
    return t_ * u_;
  }
};

template<typename T, typename U>
Mul<T, U> mul(const T& t, const U& u)
{
  return Mul<T, U>(t, u);
}

Usage: char t = 3; short u = 4; int r = mul(t, u);

Here we have two type deductions. We implicitly declare return type by usage, not exactly decltype(T*U)

Answer 3

I'm using Visual Studio 2008, so I had to come up with a non C++0x way. I ended up doing something like this.

template<typename T> struct type_precedence { static const int value = -1; };
template< > struct type_precedence<long double> { static const int value = 0; };
template< > struct type_precedence<double> { static const int value = 1; };
template< > struct type_precedence<float> { static const int value = 2; };
template< > struct type_precedence<unsigned long long> { static const int value = 3; };
template< > struct type_precedence<long long> { static const int value = 4; };
template< > struct type_precedence<unsigned long> { static const int value = 5; };
template< > struct type_precedence<long> { static const int value = 6; };
template< > struct type_precedence<unsigned int> { static const int value = 7; };
template< > struct type_precedence<int> { static const int value = 8; };
template< > struct type_precedence<unsigned short> { static const int value = 9; };
template< > struct type_precedence<short> { static const int value = 10; };
template< > struct type_precedence<unsigned char> { static const int value = 11; };
template< > struct type_precedence<char> { static const int value = 12; };
template< > struct type_precedence<bool> { static const int value = 13; };

/////////////////////////////////////////////////////////////////////////////////////////

template<typename T, typename U, bool t_precedent = ((type_precedence<T>::value) <= (type_precedence<U>::value))>
struct precedent_type { 
    typedef T t; 
};
template<typename T, typename U>
struct precedent_type<T,U,false> { 
    typedef U t;
};

/////////////////////////////////////////////////////////////////////////////////////////

template<typename T, typename U>
typename precedent_type<T,U>::t my_mul() { return T * U; }

EDIT: Here's the example - I'm actually doing this to multiply vectors. It looks something like this:

template<int N, typename T, typename U>
vec<N,typename precedent_type<T,U>::t> operator *(const vec<N,T>& v1,const vec<N,U>& v2) {
    ...
}

...

double3 = float3 * double3;
float4 = float4 * int4;
etc.

Answer 4

http://www2.research.att.com/~bs/C++0xFAQ.html#decltype