Array of pointers

Learn to create and use an array of pointers.

Array of pointers

An array of pointers is an array whose elements are pointers. In C the syntax int *arr[3]; declares an array of three int * elements.

Example: int a = 1, b = 2, c = 3; int *arr[3] = {&a, &b, &c};. Here arr[0] equals &a and arr[1] equals &b. *arr[1] gives you the value of b. Pointer arrays are common, especially for string lists like argv.

Pointer arrays for string lists

The most common use of pointer arrays is for string lists. In C a string is a char *; so a list of strings is held as an array of char *.

Example: char *names[3] = {"Ayşe", "Mehmet", "Zeynep"};. Each string literal lives in read-only memory; the array holds only pointers to them. names[0] gives the first string; names[1][0] gives the first letter of the second string. This structure is used everywhere from command line arguments to menu systems.

Exercise Print a list of names

Declare a char *names[3] array with three names (e.g. "Ayşe", "Mehmet", "Zeynep"). Use a for loop to print each name on its own line with printf.

Show example solution

#include <stdio.h>

int main(void) {
    char *names[3] = {"Ayşe", "Mehmet", "Zeynep"};
    for (int i = 0; i < 3; i++) {
        printf("%s\n", names[i]);
    }
    return 0;
}

Test yourself

Match the array of pointers parts

int *arr[3]

Array declaration
arr[0]

First pointer
*arr[1]

Pointed value
arr[i] = &x

Assignment
How do you declare an array of three strings?
- char **names[3];
- char names[3];
- char *names[3];
- char *(names);
Show answer

char *names[3]; — Char *names[3]; — three char pointers.

Print the second string of the array

char *names[3] = {"Ayşe", "Mehmet", "Zeynep"};
printf("%s\n", names[____]);

Show answer

char *names[3] = {"Ayşe", "Mehmet", "Zeynep"};
printf("%s\n", names[1]);

Next lesson: Function pointers

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