Pointer

a box saying where another box is

Pointer

Addresses

When you declare a variable, the compiler stores it at a memory location.

Given a variable named var, we can determine its memory address with &var.

#include <stdio.h>

int main() {
    int var = 5;
    printf("value of var : %d\n", var);
    printf("address of var : %p\n", &var);
    return 0;
}

If you run the code above, you'll get something like this:

value of var : 5
address of var : 0x7ffe36dfeab8

What is a Pointer?

pointer : A variable for storing addresses of a specific type. You use * to declare pointer variables.

// all of these are equivalent
int* p0;
int * p1;
int *p2;

int num;
int* addr = #
/**
 *  The integer pointer variable 'addr' contains 
 *  the address of the integer variable 'num'.
 * /

You can access the variable pointed by the pointer (called dereferencing). This is done by putting * in front of the pointer.

#include <stdio.h>

int num = 10;
int* addr = &num;

int main(void) {
    printf("value of `num` : %d\n", num);
    printf("value of `*addr` : %d\n", *addr);
    return 0;
}

Output:

value of `num` : 10
value of `*addr` : 10

Sample Code

#include <stdio.h>

int num = 1;
int* addr = &num;

int main(void) {
    printf("&num = %p \n", &num);
    printf("addr = %p \n", addr);
    printf("num = %d \n", num);
    num += 1;
    printf("*addr = %d \n", *addr);
    *addr += 1;
    printf("num = %d \n", num);
    return 0;
}

Output:

&num = 0x1086fe010 
addr = 0x1086fe010 
num = 1 
*addr = 2 
num = 3 

Pointers As Arguments of Functions

You can pass pointers as arguments of functions just like other variables.

int add(int* a, int* b) {
    return *a + *b;
}

because you pass addresses as parameters, you can change the pointed variables within functions.

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