C++ using function as parameter

Question

Suppose I have a function called

void funct2(int a) {

}


void funct(int a, (void)(*funct2)(int a)) {

 ;


}

what is the proper way to call this function? What do I need to setup to get it to work?

Answer 1

Another way to do it is using the functional library.

std::function<output (input)>

Here reads an example, where we would use funct2 inside funct:

#include <functional>
#include <iostream>

void displayMessage(int a) {
    std::cout << "Hello, your number is: " << a << '\n';
}

void printNumber(int a, std::function<void (int)> func) {
    func(a);
}

int main() {
    printNumber(42, displayMessage);
    return 0;
}

output :

Hello, your number is: 42

Answer 2

Normally, for readability's sake, you use a typedef to define the custom type like so:

typedef void (* vFunctionCall)(int args);

when defining this typedef you want the returning argument type for the function prototypes you'll be pointing to, to lead the typedef identifier (in this case the void type) and the prototype arguments to follow it (in this case "int args").

When using this typedef as an argument for another function, you would define your function like so (this typedef can be used almost exactly like any other object type):

void funct(int a, vFunctionCall funct2) { ... }

and then used like a normal function, like so:

funct2(a);

So an entire code example would look like this:

typedef void (* vFunctionCall)(int args);

void funct(int a, vFunctionCall funct2)
{
   funct2(a);
}

void otherFunct(int a)
{
   printf("%i", a);
}

int main()
{
   funct(2, (vFunctionCall)otherFunct);
   return 0;
}

and would print out:

2

Answer 3

The basic C++ way is using std::function, which acts like a "function type", for example the type int -> double in a functional language. It's not actually a type but a class template for which class instances can hold a callable function, but it's useful to think of it as used in places you would use a function type.

For example, this function takes int and returns double:

double halve(int x)
{
    return x / 2;
}

Now you can assign functions themselves (functions as first-class citizens):

std::function<double(int)> f = halve; // can store a lambda too

and you can pass it in as an argument type, or a return type. Ex:

double call_f(std::function<double(int)> f, int x)
{
    return f(x);
}

Another option is to use templates. The compiler will figure out the types and generate correctly typed code. See this comparison.

Answer 4

A more elaborate and general (not abstract) example

#include <iostream>
#include <functional>
int Add(int a, int b)
{
    return a + b;
}
int Mul(int a, int b)
{
    return a * b;
}
int Power(int a, int b)
{
    while (b > 1)
    {
        a = a * a;
        b -= 1;
    }
    return a;
}
int Calculator(std::function<int(int, int)> foo, int a, int b)
{
    return foo(a, b);
}
int main()
{
    cout << std::endl
         << "Sum: " << Calculator(Add, 1, 2);
    cout << std::endl
         << "Mul: " << Calculator(Mul, 1, 2);
    cout << std::endl
         << "Power: " << Calculator(Power, 5, 2);
    return 0;
}

Answer 5

check this

typedef void (*funct2)(int a);

void f(int a)
{
    print("some ...\n");
}

void dummy(int a, funct2 a)
{
     a(1);
}

void someOtherMehtod
{
    callback a = f;
    dummy(a)
}