Specializing C++ template based on presence/absense of a class member? - c++

Question

Consider the following:
struct A {
typedef int foo;
};
struct B {};
template<class T, bool has_foo = /* ??? */>
struct C {};
I want to specialize C so that C<A> gets one specialization and C<B> gets the other, based on the presence or absence of typename T::foo. Is this possible using type traits or some other template magic?
The problem is that everything I've tried produces a compile error when instantiating C<B> because B::foo doesn't exist. But that's what I want to test!
Edit:
I think ildjarn's answer is better, but I finally came up with the following C++11 solution. Man is it hacky, but at least it's short. :)
template<class T>
constexpr typename T::foo* has_foo(T*) {
return (typename T::foo*) 1;
}
constexpr bool has_foo(...) {
return false;
}
template<class T, bool has_foo = (bool) has_foo((T*)0)>

Answer 1

Another (C++03) approach:
template<typename T>
struct has_foo
{
private:
typedef char no;
struct yes { no m[2]; };
static T* make();
template<typename U>
static yes check(U*, typename U::foo* = 0);
static no check(...);
public:
static bool const value = sizeof(check(make())) == sizeof(yes);
};
struct A
{
typedef int foo;
};
struct B { };
template<typename T, bool HasFooB = has_foo<T>::value>
struct C
{
// T has foo
};
template<typename T>
struct C<T, false>
{
// T has no foo
};

Answer 2

Something like this might help: has_member.
typedef char (&no_tag)[1];
typedef char (&yes_tag)[2];
template< typename T > no_tag has_member_foo_helper(...);
template< typename T > yes_tag has_member_foo_helper(int, void (T::*)() = &T::foo);
template< typename T > struct has_member_foo {
BOOST_STATIC_CONSTANT(bool
, value = sizeof(has_member_foo_helper<T>(0)) == sizeof(yes_tag)
); };
template<class T, bool has_foo = has_member_foo<T>::value>
struct C {};