QUESTION 09 · Pointers and Functions

What does int (*operation)(int, int) declare?

2 marks

Answer

It declares a pointer to a function taking two int arguments and returning int.

The parentheses make it a pointer variable rather than a function returning a pointer.
A compatible function can be assigned and called through it.

Explain function pointers and callback-style use.

5 marks

Answer in brief

A function pointer identifies a function with a compatible type and allows that function to be selected or passed as a value.

Declaration and assignment

int (*operation)(int, int) = add;

Meaning

operation can identify a function receiving two ints and returning int. The function name add supplies a suitable function pointer in this expression. Calling operation(4, 6) or (*operation)(4, 6) invokes the selected function. The function must have a compatible type; an arbitrary cast does not make an incompatible call safe.

Callback-style example

apply receives a function pointer named op along with a and b. It returns op(a, b). main supplies add for one call and subtract for another. Thus apply keeps the general calling step while the caller chooses the operation. The output is 10 for addition and -2 for subtraction.

Uses

Select an arithmetic operation or menu action.
Let a general algorithm call a caller-supplied comparison or processing function.
Keep a table of compatible operations when that makes the design clearer.

Precautions

Do not call a null or invalid function pointer. An object pointer and a function pointer are not interchangeable portable storage formats. The sample always passes real functions with exactly matching signatures.

Complexity

An indirect call does not change the algorithmic growth of the selected function. With the fixed-size addition or subtraction shown, each application takes O(1) time and O(1) extra space. More complex callbacks can perform more work.

C program and output

Download .c Open compiler

/* CodeBhavya | PPS AR25 | Learning example */
#include <stdio.h>
int add(int a, int b) { return a + b; }
int subtract(int a, int b) { return a - b; }
int apply(int (*op)(int, int), int a, int b)
{
    return op(a, b);
}

int main(void)
{
    printf("Add = %d\n", apply(add, 4, 6));
    printf("Subtract = %d\n", apply(subtract, 4, 6));
    return 0;
}

Sample input and output

This example uses values already written in the program.

Output

Add = 10
Subtract = -2