std::enable_if specialisation failing

Question

I've been messing around with enable_if, and I seem to have stumbled upon some inconsistent behaviour. This is in VS2010. I've reduced it to the following sample.

#include <type_traits>

using namespace std;

// enable_if in initial template definition
template <class T, class Enable = enable_if<true>> struct foo {};

foo<int> a; //OK

// explicit specialisation
template <class T, class Enable = void> struct bar;
template <class T> struct bar<T, void> {};

bar<int> b; //OK

// enable_if based specialisation
template <class T, class Enable = void> struct baz;
template <class T> struct baz<T, std::enable_if<true>> {};

baz<int> c; //error C2079: 'c' uses undefined struct 'baz<T>'

Is this a bug in the code or the compiler?

Answer 1

std::enable_if<true> should be typename std::enable_if<true>::type .

std::enable_if<true> always names a type (as does std::enable_if<false> ). In order to get substitution to fail when the condition is false you need to use the type nested typedef, which is only defined if the condition is true.

Answer 2

Your problem has very little to do with enable_if

// you declare a structure named baz which takes 2 template parameters, with void
// as the default value of the second one.
template <class T, class Enable = void> struct baz;
// Partial specialization for a baz with T and an std::enable_if<true>
template <class T> struct baz<T, std::enable_if<true>> {};

// Declare a baz<int, void> (remember the default parameter?):
baz<int> c; //error C2079: 'c' uses undefined struct 'baz<T>'

baz<int, void> has an incomplete type at that point. The same problem will occur without enable_if :

template <class T, class U = void>
struct S;

template <class T>
struct S<T, int>
{ };

S<double> s;

And, as James said, you're using enable_if incorrectly. Boost's documentation for enable_if does a great job explaining it.