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Passing Pointers to Functions in C

When you pass a pointer to a function, you are sending the address of a variable. This allows the function to directly read or modify the original variable — unlike normal pass-by-value where only a copy is sent.

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Syntax

void function_name(int *ptr) {
    // use *ptr to access or modify the original value
}

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Example: Modify a Variable via Pointer

#include <stdio.h>

void changeValue(int *p) {
    *p = 100;   // dereference to change original
}

int main() {
    int x = 10;
    printf("Before: x = %d\n", x);

    changeValue(&x);   // pass address of x

    printf("After : x = %d\n", x);
    return 0;
}
Output: 
Before: x = 10
After : x = 100

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Example: Swap Two Numbers

Without pointers, swapping inside a function has no effect on the originals. With pointers: 
#include <stdio.h>

void swap(int *a, int *b) {
    int temp = *a;
    *a = *b;
    *b = temp;
}

int main() {
    int x = 5, y = 10;
    printf("Before: x=%d, y=%d\n", x, y);
    swap(&x, &y);
    printf("After : x=%d, y=%d\n", x, y);
    return 0;
}
Output: 
Before: x=5, y=10
After : x=10, y=5

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Pass by Value vs Pass by Reference

void noEffect(int n) {
    n = 999;           // only changes local copy
}

void withEffect(int *n) {
    *n = 999;          // changes original
}

int main() {
    int a = 10, b = 10;
    noEffect(a);       // a is still 10
    withEffect(&b);    // b becomes 999
    printf("%d %d", a, b);   // Output: 10 999
}

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Key Points

  • Pass &variable at the call site to send the address
  • Use *ptr inside the function to access or modify the value
  • This is how C achieves pass-by-reference behaviour
  • Especially useful for functions that need to return multiple values