In namespace std, of the gnu stdc++ header ostream, this is the definition of std::endl:
template<typename _CharT, typename _Traits>
inline basic_ostream<_CharT, _Traits>&
endl(basic_ostream<_CharT, _Traits>& __os)
{ return flush(__os.put(__os.widen('\n'))); }
Why is it possible to write std::cout << std::endl; where endl is used without template parameters?
std::basic_ostream<CharT,Traits>::operator<< taking manipulators as function pointer. Template argument deduction is performed when being passed template like std::endl, thus template arguments don't need to be specified explicitly.
When possible, the compiler will deduce the missing template arguments from the function arguments. This occurs when a function call is attempted and when an address of a function template is taken.
This mechanism makes it possible to use template operators, since there is no syntax to specify template arguments for an operator other than by re-writing it as a function call expression.
#include <iostream> int main() { std::cout << "Hello, world" << std::endl; // operator<< is looked up via ADL as std::operator<<, // then deduced to operator<<<char, std::char_traits<char>> both times // std::endl is deduced to &std::endl<char, std::char_traits<char>> }
Given std::cout << std::endl;, the parameter type of the appropriate operator<< is std::basic_ostream<char,std::char_traits<char>>& (*func)(std::basic_ostream<char,std::char_traits<char>>&), then the template arguments of std::endl would be deduced as char and std::char_traits<char>.