How to check if two template parameters are exactly the same?

Question

How do I modify the following function template so that it returns 42 if template parameters T and U are exactly the same type?

template<typename T,typename U>
int Foo()
{
  return 0;
}

Answer 1

Using std::is_same can provide the desired behaviour:

#include <type_traits>

template<typename T,typename U>
int Foo()
{
    return std::is_same<T, U>::value ? 42 : 0;
}

Answer 2

An idiomatic way is to delegate the work to a helper function object in a detail namespace that you can partially specialize for the case where T is the same as U (or for any other compile-time pattern that you can use in class templates).

namespace detail {

template<typename T, typename U>
struct foo
{
     int operator()() const
     {
         return 0;
     }
};

template<typename T>
struct foo<T, T>
{
     int operator()() const
     {
         return 42;
     }
};

} // namespace detail 

template<typename T, typename U>
int Foo()
{
     return detail::foo<T, U>()();
}

For functions which also have deducible arguments (e.g. a Foo(T x, U y) would) this combines the power of argument deduction of function templates, and the specialization capabilities of class templates, without users every being the wiser (well, you need the convention that they do not call anything from namespace detail directly)