Using function as the argument of another function

Asked Apr 27 '20 at 22:17

Active Apr 27 '20 at 22:23

Viewed 32 times

I have on function that has takes another function as the argument:

#include <iostream>
using namespace std;

int addition (int a, int b)
{ return (a+b); }

int subtraction (int a, int b)
{ return (a-b); }

int operation (int x, int y, int functocall(int,int))
{
  int g;
  g = functocall(x,y);
  return (g);
}

When i use it in the main function:

int main ()
{
  int m,n;
  int (*minus)(int,int) = subtraction;

  m = operation (7, 5, addition);
  n = operation (20, m, *minus);
  cout <<n;
  return 0;
}

Other ways to use addition and minus in operation:

  m = operation (7, 5, &addition);
  n = operation (20, m, *minus);

or:

  m = operation (7, 5, *addition);
  n = operation (20, m, minus);

All of them have the same result: 8 in this case. Are there any differences between them ? What should i use ?

edited Apr 27 '20 at 22:23

asked Apr 27 '20 at 22:17

Hitokiri

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functions are their own pointers. It's super weird. – Mooing Duck Apr 27 '20 at 22:25
From https://timsong-cpp.github.io/cppwp/n3337/conv.func *An lvalue of function type `T` can be converted to a prvalue of type “pointer to `T`.” The result is a pointer to the function.* – R Sahu Apr 27 '20 at 22:29
m = operation (7, 5, addition); is more user friendly. IMO – shanker861 Apr 27 '20 at 22:31

Using function as the argument of another function

0 Answers0