Thread: Can I take function names as arguments without using templates?

STL supposedly does this for functors right, algorithms use identifiers of functions/lambda expressions that return a boolean and get passed two arguments.

I want to do something similar.
I could take a function identifier as argument for another function using templates like so:

Code:

template <typename T>
void call_function(T func_name) {
    func_name();
    std::cout << "\n\n" << typeid(T).name();
// type = bool (__cdecl*)(void)
}

void print() {
    std::cout << "hey";
}

int main()
{

    call_function(print);

    std::cin.get();
}

But how do I do this without templates? Suppose I want to allow only void functions that take no values?

From that cout statement I got to know that T evaluated to bool (__cdecl*)(void)

But I can't use bool (__cdecl*)(void) as a type for an argument.

Use a function pointer parameter, e.g.,

Code:

void call_function(void (*func_name)())

A typedef for the function pointer type may be useful.