Can the types of parameters in template functions be inferred?

Question

I'm writing some template functions in C++, but I'm not sure if it's possible to define a template function that infers the types of its parameters.

I tried to define a template with inferred parameter types, but this example won't compile:

template <auto>   
auto print_stuff(auto x, auto y) 
{ 
    std::cout << x << std::endl;
    std::cout << y << std::endl;
}

It works when I give a unique name to each parameter type, but this seems somewhat redundant:

#include <iostream> 
#include <string>

template <class Redundant_1,class Redundant_2>   
auto print_stuff(Redundant_1 x, Redundant_2 y) 
{ 
    std::cout << x << std::endl;
    std::cout << y << std::endl;
}

int main() 
{ 
    print_stuff(3,"Hello!");
    return 0; 
}

Is it possible to define a template with inferred parameter types instead of giving each type a unique name?

Answer 1

You can dispense with the template-header and names for the parameter-types if your compiler supports concepts, which isn't generally enabled even if asking for experimental C++2a mode.
On gcc for example, it must be separately enabled with -fconcepts.

See live on coliru.

#include <iostream> 
#include <string>

auto print_stuff(auto x, auto y) 
{ 
    std::cout << x << std::endl;
    std::cout << y << std::endl;
}

int main() 
{ 
    print_stuff(3,"Hello!");
    return 0; 
}

As an aside, avoid std::endl and use std::flush in the rare cases you cannot avoid costly manual flushing. Also, return 0; is implicit for main().