Question

This in VS2010sp1 doesn’t compile (it does compile with gcc 4.6 though):

template<class T>
struct Upcast;

template<>
struct Upcast<signed char>
{
    typedef signed short type;
};

template<>
struct Upcast<char>
{
    typedef typename std::conditional<std::is_signed<char>::value,short, unsigned short>::type type;
};

int main()
{
    Upcast<char>::type a;
    return 0;
}

Error from VS:

Error   1   error C2899: typename cannot be used outside a template declaration

Which team is right? VS or gcc?

Answer 1

VS is right on C++03. GCC is right on C++0x.

Now it may be sensible for GCC to also allow this in C++03 mode (there are many things real compilers don’t diagnose in C++03 mode that are actually only valid in C++0x), and it may as-well sensible for VS to reject it in C++03 mode.

It doesn’t matter anymore whether or not a use of typename QualifiedName happens in a template or not, in C++0x. That is, the following is perfectly legal for C++0x:

#include<vector>

int main() {
  typename std::vector<int> v;
}

In C++03, typename could only be used inside of a template. And the explicit specialization in your code is not a template. There are no template<typename T ...> clauses (all parameters in your code are fixed).