ambiguous template wierdness

Question

I have the following code (sorry for the large code chunk, but I could not narrow it down any more)

template <bool B>
struct enable_if_c {
      typedef void type;
};

template <>
struct enable_if_c<false> {};

template <class Cond>
struct enable_if : public enable_if_c<Cond::value> {};

template <typename X>
struct Base { enum { value = 1 }; };

template <typename X, typename Y=Base<X>, typename Z=void>
struct Foo;

template <typename X>
struct Foo<X, Base<X>, void> { enum { value = 0 }; };

template <typename X, typename Y>
struct Foo<X, Y, typename enable_if<Y>::type > { enum { value = 1 }; };

int main(int, char**) {
        Foo<int> foo;
}

but it fails to compile with gcc (v4.3) with

foo.cc: In function ‘int main(int, char**)’:
foo.cc:33: error: ambiguous class template instantiation for ‘struct Foo<int, Base, void>’
foo.cc:24: error: candidates are: struct Foo<X, Base<X>, void>
foo.cc:27: error:                 struct Foo<X, Y, typename enable_if<Y>::type>
foo.cc:33: error: aggregate ‘Foo<int, Base<int>, void> foo’ has incomplete type and cannot be defined

ok, so it's ambiguous. but I was'nt expecting it to be a problem as when using specialization it will almost always be some ambiguity. However this error is only triggered when using the class with enable_if<...>, if I replace it with a class like the following there is no problem.

template <typename X, typename Y>
struct Foo<X, Y, void > { enum { value = 2 }; };

why does this class not cause a ambiguity while the other do? isn't the two the same thing for classes with a true ::value? anyway, any hints as to what im doing wrong is appreciated.

Thanks for the answers, my real problem (to get the compiler to select my first specialization) was solved by replacing struct Foo<X, Base<X>, void> with struct Foo<X, Base<X>, typename enable_if< Base<X> >::type >. Which seems to work the way I want.

Answer 1

aren't you missing a

<

symbol ?

Answer 2

I think you are missing a '<', a template should look like:

template< typename T >
struct myStruct
{};

//OR

template< class T >
struct myStruct
{};

Answer 3

The gist of your question is that you have:

template <typename X, typename Y, typename Z>
struct Foo {};

template <typename X>
struct Foo<X, Base<X>, void> {};                   // #1

template <typename X, typename Y>
struct Foo<X, Y, typename whatever<Y>::type> {};   // #2

and you're trying to match it to

Foo<int, Base<int>, void>

Obviously, both specializations match (the first with X = int, the second with X = int, Y = Base<int>).

According to the standard, section 14.5.4, if there are more matching specializations, a partial ordering (as defined in 14.5.5.2) among them is constructed and the most specialized one is used. In your case, however, neither one is more specialized than the other. (Simply put, a template is more specialized than another, if you can replace each type parameter of the latter template with some type and in result get the signature of the former. Also, if you have whatever<Y>::type and you replace Y with Base<X> you get whatever<Base<X> >::type not void, i.e. there is not processing performed.)

If you replace #2 with

template <typename X, typename Y>
struct Foo<X, Y, void > {};                        // #3

then the candidate set again contains both templates, however, #1 is more specialized then #3 and as such is selected.