Why doesn't C++ find template function?

Question

Why do I get the compile error no matching function for call to `f( __gnu_cxx::__normal_iterator > >)'?

#include <vector>

template<typename T>
void f(const typename std::vector<T>::iterator &) {}

void g() {
  std::vector<int> v;
  f<int>(v.end());  // Compiles.
  f(v.end());  // Doesn't compile, gcc 4.3 can't find any match.
}

Ultimately I want to write a function which takes only a vector iterator, and fails to compile (with a meaningful error) for anything else. So template<typename T>void f(const T&) {} is not a good solution, because it compiles for other types as well.

Answer 1

You cannot deduce a template argument from a nested type. Think of, e.g., std::vector<T>::size_type which is always std::size_t: how would the compiler resolve the ambiguity? I realize that's not quite the case in your example but the same principle applies. For example, the iterator type of std::vector<T> can be T* which can also be the iterator type of std::array<T, N>.

Answer 2

G++ 4.8 gives a more complete message: http://ideone.com/ekN3xs

note:   template argument deduction/substitution failed:
note:   couldn't deduce template parameter ‘T’

f doesn't directly takes a T (like in "const T&") or a type where T is clear (like in "const std::vector<T>&") but a nested dependent type (here std::vector<T>::iterator) so the template type T cannot be automatically deduced from the argument.

Edit: Dietmar Kühl's answer gives a good rationale example.

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For your "Ultimately" part, check How to check whether a type is std::vector::iterator at compile time? (the accepted answer uses some C++11 types, but you can use C++03 equivalents, e.g. from Boost)