address of c++ template function

Question

Why does this fail to compile? (g++-4.5)

template < typename U >
static void h () {
}

int main () {
  auto p = &h<int>; // error: p has incomplete type
}

EDIT: Here is a work-around:

template < typename U >
static void h () {
}

int main () {
  typedef decltype (&h<int>) D;
  D p = &h<int>; // works
}

Answer 1

It does not compile because type of 'p' is not known to the compiler which is a must in C++ unlike some other languages.

Try

template < typename U > 
static void h () { 
} 

int main () { 
  auto void (*p)() = &h<int>; 
} 

Answer 2

In C++0x this is guaranteed to work. However in C++03 this wasn't working (the initializer part, that is) and some compilers apparently don't support it yet.

Furthermore, I remember that the C++0x wording is not clear what happens with &h<int> when it is an argument to a function template and the corresponding parameter is deduced (this is what auto is translated to, conceptionally). The intention is, however, that it is valid. See this defect report where they designed the wording, the example by "Nico Josuttis" and their final example.

There is another rule that the wording enforces but compilers are not correctly implementing. For example, see this clang PR.

Answer 3

Try

 auto p  = static_cast<void(*)()>(& h<int>);

Because gcc treats templated function as overloaded one. From the gcc's point of view it's like you would have h(int param) and h(float param) - which one the compiler has to choose?

edit: this compiles on gcc so please give me a reason why downvoted. edit2: I didn't receive an answer so I assume that's because you're an idiot.