c++ - std library map and template functions -

this question has answer here:

• where , why have put “template” , “typename” keywords? 5 answers

i don't understand why code

#include <map>  template<typename t, typename u> std::ostream& operator<<(std::ostream& o,         const std::map<t,u>& input) {     (std::map<typename t,typename u>::iterator it=input.begin(); it!=input.end(); ++it)     {         o << it->first << " => " << it->second << '\n';     }     return o; } 

returns compilation error:

error: wrong number of template arguments (1, should 4) error: provided ‘template<class _key, class _tp, class _compare, class _alloc> class std::map’ 

can me please??

you should write typename before iterator declaration , use const_iterator:

for (typename std::map<t,u>::const_iterator it=input.begin(); it!=input.end(); ++it

the argument of operator << requires const objects. elements of map must const. achieve 1 uses const_iterator. typename in declaration of iterator required indicate following expression nested template class depending on types t , u.

see question: making sense of arguments function template