Mixing template specialisation and enable_if in a function template with only a templated return type

Question

I have the following code that fails to compile on VC2010:

#include <type_traits>

using namespace std;

template <class C>
typename enable_if<true, C>::type
foo()
{ return C(); }

template <>
bool
foo()
{ return true; } // error C2785: 'enable_if<true,_Type>::type foo(void)' 
                 // and 'bool foo(void)' have different return types

int main()
{
    auto a = foo<int>();
    auto b = foo<bool>();
}

The error message seems erroneous, as the first version of foo() seems functionally identical to template <class C> C foo(); which happily compiles.

Is there a way to mix-and-match enable-if'd function templates and explicit template specialisations?

Answer 1

A function template specialization is (thankfully!) not required to return the same type as the non-specialized template, so this is not the issue here.

In fact, enable_if has nothing to do with your error, your code is simply missing the template arguments list in the specialization:

template <>
bool foo<bool>()
{ return true; }

On a sidenote, why are you using enable_if if the condition is always true? (I guess this is not the case in your real code, but I just want to be sure :)!)

Answer 2

The problem is just the syntax of the full specialization. It should be:

template <> bool foo<bool>() { return true; }
                    ^^^^^^