SFINAE working in return type but not as template parameter

Question

I already used the SFINAE idiom quite a few times and I got used to put my std::enable_if<> in template parameters rather than in return types. However, I

Answer 1

The = ... of the template just gives a default parameter. This is not part of the actual signature which looks like

template
auto foo(T a);

for both functions.

Depending on your needs, the most generic solution for this problem is using tag dispatching.

struct integral_tag { typedef integral_tag category; };
struct floating_tag { typedef floating_tag category; };

template  struct foo_tag
: std::conditional::value, integral_tag,
                    typename std::conditional::value, floating_tag,
                                               std::false_type>::type>::type {};

template
T foo_impl(T a, integral_tag) { return a; }

template
T foo_impl(T a, floating_tag) { return a; }

template 
T foo(T a)
{
  static_assert(!std::is_base_of >::value,
                 "T must be either floating point or integral");
  return foo_impl(a, typename foo_tag::category{});
}

struct bigint {};
template<> struct foo_tag : integral_tag {};

int main()
{
  //foo("x"); // produces a nice error message
  foo(1);
  foo(1.5);
  foo(bigint{});
}