The & address operator

Learn to take a variable's address with &.

The & address operator

The & operator gives the memory address of a variable. This address can be assigned to a pointer so that the pointer refers to that variable.

int x = 10;
int *ptr = &x;
printf("%p\n", (void*)ptr);  /* prints the address */

The address is the variable's location in memory; it may differ between runs. The %p format specifier is used to print addresses. Without an address, you cannot tell a pointer which variable to track.

The & operator with scanf

The scanf function asks for a variable's address to put input into it. That is why & is used.

int age;
printf("Age: ");
scanf("%d", &age);  /* writes to age's address */
printf("%d\n", age);

Without & before age, scanf does not work; the function would not know where to write. This is an everyday use of addresses.

Exercise Take the addresses of two variables

Define int a = 3, b = 5;. Take each address with int *pa = &a, *pb = &b;. Print the values with printf("%d %d\n", a, b);. Output should be 3 5.

Show example solution

#include <stdio.h>

int main(void) {
    int a = 3, b = 5;
    int *pa = &a, *pb = &b;
    printf("%d %d\n", a, b);
    return 0;
}

Test yourself

Take an address and assign to a pointer

int x = 5;
int *ptr = ____;  /* ptr holds x's address */

Show answer

int x = 5;
int *ptr = &x;  /* ptr holds x's address */

What does & do in scanf("%d", &x)?
- Gives the address of x
- Resets x
- Prints x
- Gives the value of x
Show answer

Gives the address of x — &x gives the address of x; scanf writes to that address.
Match address and pointer

&

address operator
int *

int pointer type
%p

address format
scanf

writes to address

Next lesson: Pointer dereference

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