function pointer that takes and returns a pointer

Question

I'm trying to get this code to run..

#include <stdio.h> 

int *myMethod(int *a) 
{ 
   printf("Hello");
   return a; 
} 

int main() 
{ 
    // my_ptr is a pointer to function myMethod() 
    int *my_ptr(int *) = myMethod; 

    // Invoking myMethod() using my_ptr 

    int a = 5;
    int *p = (*my_ptr)(&a); 
    printf("Bye %d\n", *p);

    return 0; 
}

I thought my syntax for a function pointer, my_ptr, would be correct where it could take an int pointer as it's parameter and return an int pointer but when I compile it, I get the error that:

 error: function 'my_ptr' is initialized like a variable
 int *my_ptr(int *) = myMethod;

Could someone explain the error/issue? Thanks!

Start easier, use a function returning and taking an integer, not a pointer, and you will see wher your problem lies. — Ulrich Eckhardt, Nov 04 '18 at 07:15

score 1 · Answer 1 · answered Nov 04 '18 at 06:55

1

int* my_ptr(int*) is the prototype of a function. You want a function pointer: int* (*my_ptr)(int*)

answered Nov 04 '18 at 06:55

Nelfeal

12,593
1
20
39

score 1 · Answer 2 · answered Nov 04 '18 at 07:01

1

Change pointer to function myMethod to this:

int *(*my_ptr)(int *) = &myMethod;

answered Nov 04 '18 at 07:01

Farhad Mortezapour

233
1
8

score 0 · Answer 3 · answered Nov 04 '18 at 07:00

You should use int *(*my_ptr)(int *) = myMethod; not int *my_ptr(int *) = myMethod;

The following code could work:

#include <stdio.h>

int *myMethod(int *a)
{
   printf("Hello");
   return a;
}

int main()
{
    int *(*my_ptr)(int *) = myMethod;

    int a = 5;
    int *p = (*my_ptr)(&a);
    printf("Bye %d\n", *p);

    return 0;
}

function pointer that takes and returns a pointer

3 Answers3