Templates for cout input - c++

Question

I want to create a function that takes in anything that the << operator for std::cout can handle. I have an example that breaks.
#include <iostream>
template <typename T>
void my_print(const T &t) {
std::cout << t;
}
int main() {
my_print("hello\n"); // works
my_print(4); // works
my_print(std::endl); // compiler error
return 0;
}
It also fails if I change to void my_print(T t). The compiler error is
error: no matching function for call to 'my_print(<unresolved overloaded function type>)'
note: candidate is
note: template<class T> void my_print(const T&)
Why can't the compiler resolve it when it sees that the argument t is being put into cout?
Is there any good way to fix this or do I have to manually provide the additional << cases, e.g. void my_print(ostream& (*pf)(ostream&));
EDIT: I know endl is a function. Is the answer then that function types are not accepted as templates? Like I can't have [T = ostream& (*)(ostream&)]?

Answer 1

std::endl is actually a function template. You can read the full documentation here or here. It is defined as:
template< class CharT, class Traits >
std::basic_ostream<charT,traits>& endl( std::basic_ostream<CharT, Traits>& os );
Edit: You can achieve what you want using this solution (which I vaguely adapted from here)
#include <iostream>
// handles the other types
template <typename T>
void my_print(const T &t) {
std::cout << t;
}
// alias a few things to make the prototypes readable
typedef std::basic_ostream<char, std::char_traits<char> > CoutType;
typedef CoutType& (*StandardEndLine)(CoutType&);
int main() {
my_print("hello\n"); // works
my_print(4); // works
my_print((StandardEndLine)std::endl); // <- NOTE: there is an explicit cast
return 0;
}