std::endl

[Alby94]

Why this code report me this error??
Overload function with no contextual type information.

It isn't a function's overload...

prototype: ostream& endl (ostream& os);

(*endl)(cout);

[coder777]

Actually endl is a function. the ostream operator << accepts a function of that type.

Read this:
http://www.cplusplus.com/reference/ostream/ostream/operator%3C%3C/

[JLBorges]

Prototype:

template< class CHAR_TYPE, class TRAITS_TYPE >
std::basic_ostream< CHAR_TYPE, TRAITS_TYPE >& std::endl( std::basic_ostream< CHAR_TYPE, CHAR_TYPE >& ) ;

Either: std::endl( std::cout ) ; // types are deduced

Or: ( *std::endl< char, std::char_traits<char> > )(std::cout) ;