std::enable_if specialisation failing

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 开发者_如何学运维or the compiler?

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.

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.