Introduction

Pointers are found when working with lower level programming languages(languages 'closer' to machine code).

In the following examples you'll use a main.c which is compiled & then run.

gcc main.c -o main
./main

Pass a variable

What do you expect x to print?

#include <stdio.h>

void update(int p) {
    p = 42;
}

int main() {
    int x = 10;
    update(x);
    printf("%d\n", x); // 10
    return 0;
}

We invoke update() which reassigns it's local p and then print the value of x after update()() returns. 10 prints because x's value wasn't changed

Pass a reference/pointer

How might we change the value of x from update()?

#include <stdio.h>

void update(int *p) {
    *p = 42;
}

int main() {
    int x = 10;
    update(&x);
    printf("%d\n", x); // 42
    return 0;
}

Copying

If we wanted to swap two variables we could use a struct to store the values we want to swap. After we invoke swap() we then use the it's fields/members a & b to update x & y.

#include <stdio.h>

typedef struct {
    int a;
    int b;
} Pair;

Pair swap(int a, int b) {
    Pair p;
    p.a = b;
    p.b = a;
    return p;
}

int main() {
    int x = 5, y = 10;
    printf("Before swap: x = %d, y = %d\n", x, y);

    Pair result = swap(x, y);
    x = result.a;
    y = result.b;

    printf("After swap: x = %d, y = %d\n", x, y);
    return 0;
}

Swapping Values

A better approach would be to swap in place using references(&) & pointers(*).

We pass the references of x & y to swap() and then update it using the pointer syntax, *.

#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 swap: x = %d, y = %d\n", x, y);

    swap(&x, &y);

    printf("After swap: x = %d, y = %d\n", x, y);
    return 0;
}

Conclusion

By passing