Pointer dereference
- C course
- Beginner
- Lesson 73/116
- 5 min
Learn to access the value a pointer points to with *.
Pointer dereference with *
The * operator reads (or writes) the value at the address a pointer holds. This is called dereferencing.
int x = 10;
int *ptr = &x;
printf("%d\n", *ptr); /* prints 10 */
*ptr = 20; /* x is now 20 */*ptr works for both reading and writing. Assignment changes the value at the pointed address; it changes x, not ptr.
Changing a value via dereference
The assignment *ptr = value; writes to the address the pointer holds. So the variable ptr points to changes.
int x = 5;
int *ptr = &x;
*ptr = 99; /* x is now 99 */
printf("%d\n", x); /* prints 99 */This is the basis for functions changing original variables. The function takes a pointer, writes to the address with *.
Practice Change a value via dereference and print
Define int x = 5;. Bind with int *ptr = &x;. Set x to 30 with *ptr = 30;. Print with printf("%d\n", x);. Output should be 30.
Starter code
#include <stdio.h>
int main(void) {
int x = 5;
/* bind ptr to x */
/* set x to 30 via dereference */
/* print x with printf %d */
return 0;
}Sample solution
#include <stdio.h>
int main(void) {
int x = 5;
int *ptr = &x;
*ptr = 30;
printf("%d\n", x);
return 0;
}Expected output
30
Questions and answers
Review the lesson questions together with their answers. To test yourself, solve the lesson interactively.
1 Match pointer operators
| Term | Meaning |
|---|---|
& |
takes address |
* |
dereferences |
int * |
pointer type |
NULL |
null pointer |
2 Read a value via dereference
int x = 7;
int *ptr = &x;
printf("%d\n", ____); /* prints 7 */int x = 7;
int *ptr = &x;
printf("%d\n", *ptr); /* prints 7 */Explanation: *ptr gives the value ptr points to.
3 What does `*ptr = 99;` change?
- ptr itself
- The variable ptr points to Correct answer
- Nothing
- Only ptr's address
Explanation: *ptr assignment writes to the address ptr points to; that variable changes.