How to make a default argument to parameter which takes function address?

Question

How would I make a function that takes address of another function as a parameter so that passing it is not necessary.

My first solution was this:

void empty_func() {}

void func(void (*additional_func) () = &empty_func)
{
    additional_func();
    // some other code...
}

This solution works.

However, I was wondering is there some more convenient way to make passing additional_function optional, so I do not have to make a new function (empty_func()) and passing it as a default argument?

Answer 1

As function pointers are also a pointer and null-equivalent for pointer types are nullptr so you can assign the value to this default parameter as nullptr.

Also, don't miss forget to add nullptr check before using it.

void func(void (*additional_func) () = nullptr )
{
    if(additional_func)
    {
        additional_func();
    }

    //some other code...
}

Also, it is more convenient to define an alias for this function-pointer as use like following:

using TPFunc = void (*) ();

void func( TPFunc additional_func = nullptr )
{
    if(additional_func)
    {
        additional_func();
    }

    //some other code...
}

But it is best to use features from functional header file instead of these raw-function-pointers. Following is an example to achieve this using functional header file:

#include <iostream>
#include <functional>
 

using TPFunc = std::function<void()>;

void func( TPFunc additional_func = nullptr )
{
    if(additional_func)
    {
        additional_func();
    }
    else
    {
        std::cout<< "Function pointer is `nullptr`\n";
    }

    //some other code...
}


void fun1()
{
    std::cout << "In 'void fun1()'\n";
}


void fun2()
{
    std::cout << "In 'void fun2()'\n";
}

int main()
{
    func();
    func(fun1);
    func(fun2);
    std::cout<< "Done";
    return 0;
}

Output:

Function pointer is `nullptr`
In 'void fun1()'
In 'void fun2()'
Done