Passing Function name as arguments to a function

Question

Is it possible to pass the name of a function(say A) as an argument to another function (say B), and then call function A from function B. That is the function name will be stored in a variable in B, and using it call the function whose name is in the variable. For Example in C++ sort function the first and second arguments are iterators, but the third argument is the name of a function.

Answer 1

You can introduce a template parameter (here Pred) for "anything that is callable with two parameters":

template <typename Iter, typename Pred>
void mysort(Iter begin, Iter end, Pred predicate)
{
    --end;
    // ...
        if (predicate(*begin, *end))
        {
            // ...
        }
    // ...
}

Then you can pass either good old C function pointers or C++ function objects:

bool function(int x, int y)
{
    return x < y;
}

struct function_object
{
    bool operator()(int x, int y)
    {
        return x < y;
    }
};

int main()
{
    int numbers[] = {543, 6542654, 432514, 54, 45, 243};
    mysort(numbers + 0, numbers + 6, &function);
    mysort(numbers + 0, numbers + 6, function_object());
}

As you can see, a function object is an object of a class that overloads operator() appropriately.

Answer 2

You should read up on Function Pointers.

Answer 3

simple example with Function Pointers:

#include <iostream>

int apply(int (*fun)(int,int), int a, int b) {
    return (*fun)(a,b);
}

int add(int a, int b) {return a + b;}
int multiply(int a, int b) {return a * b;}

int main(int argc, const char* argv[]) {
    int added = apply(add, 2, 4);
    int multiplied = apply(multiply, 2, 4);

    std::cout << "added result: " << added << std::endl;
    std::cout << "multiplied result: " << multiplied << std::endl;
}

output:

added result: 6
multiplied result: 8