#include <iostream> template<typename... Args> void print(Args const&... args) { (std::cout << ... << args); } int main() { std::cout << 1 << 2 << 3 << std::endl; // ok print(1, 2, 3); // ok print(1, 2, 3, std::endl); // error! How to make it work? } See online demo
How to pass a function template as a template argument?
You will have the same issue with other io manipulators that typically are functions that take the stream as parameter, when they are templates. Though you can wrap them in a non-template callable:
#include <iostream> template<typename... Args> void print(Args const&... args) { (std::cout << ... << args); } int main() { std::cout << 1 << 2 << 3 << std::endl; // ok print(1, 2, 3); // ok print(1, 2, 3, [](std::ostream& o) -> std::ostream&{ o << std::endl; return o; }); // no error! } 123 123123 The syntax is rather heavy so you might want to use a helper type, though I'll leave it to you to write that (just joking, I don't think it is trivial, but I might give it a try later ;). After pondering about it for a while, I am almost certain that there are only the two alternatives: Instantiate the function (see other answer), or wrap the call inside a lambda, unless you want to write a wrapper for each single io manipulator of course.
Here a way:
print(1, 2, 3, std::endl<char, std::char_traits<char>>); Consider using '\n' instead.
'\n' does not flush buffer, but std::endl does!. What if OP wants that! Also have a look: Can I take the address of a function defined in standard library?. Therefore, packing into a lambda is the better way.
You cannot take address of most standard functions (see can-i-take-the-address-of-a-function-defined-in-standard-library).
Fortunately, io-manipulator is part of the exception (See Addressable_functions).
std::endl is a template function, so you would have to select the correct overload.
using print_manip_t = std::ostream& (*) (std::ostream&); print(1, 2, 3, print_manip_t{std::endl}); print(1, 2, 3, static_cast<print_manip_t>(std::endl)); print(1, 2, 3, static_cast<std::ostream& (*) (std::ostream&)>(std::endl)); else you have to specify which one you want
print(1, 2, 3, std::endl<char, std::char_traits<char>>); or wrap it
print(1, 2, 3, [](std::ostream& o) -> std::ostream&{ return o << std::endl; }); 
print(1, 2, 3, std::endl<char, std::char_traits<char>>);, but you probably want a way to not have to write all that extra stuff.'\n'(instead ofstd::endl).\n,print(1, 2, 3, '\n');unless you absolutely must flush the stream too.