Why doesn't C++ find template function? -

why compile error no matching function call `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 match. } 

ultimately want write function takes vector iterator, , fails compile (with meaningful error) else. template<typename t>void f(const t&) {} not solution, because compiles other types well.

g++ 4.8 gives 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 t (like in "const t&") or type t clear (like in "const std::vector<t>&") nested dependent type (here std::vector<t>::iterator) template type t cannot automatically deduced argument.

edit: dietmar kühl's answer gives rationale example.

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for "ultimately" part, check how check whether type std::vector::iterator @ compile time? (the accepted answer uses c++11 types, can use c++03 equivalents, e.g. boost)