Array size

Learn to compute the array size with sizeof.

Array size with sizeof

C has no operator that directly gives array length. But sizeof tells you both the total bytes of the array and the bytes of one element; the division gives the length.

int scores[5] = {85, 92, 78, 90, 88};
int size = sizeof(scores) / sizeof(scores[0]);
printf("%d\n", size);  // 5

Here sizeof(scores) is 20 bytes (5 × 4), sizeof(scores[0]) is 4 bytes. The division gives 5. This idiom is useful for loop bounds.

sizeof as loop bound

Using the sizeof idiom for loop bounds is safer than hard-coding the size. If you change the size later, the loop adapts automatically.

int scores[5] = {85, 92, 78, 90, 88};
int n = sizeof(scores) / sizeof(scores[0]);
for (int i = 0; i < n; i++) {
    printf("%d\n", scores[i]);
}

This style makes maintenance easier. If the array grows to 10, the loop runs 10 times. Hard-coding 5 would risk forgetting the update.

Exercise Print the array length

Write a C program. Declare int scores[5] = {85, 92, 78, 90, 88};. Compute the length with int n = sizeof(scores) / sizeof(scores[0]);. Print with printf("%d\n", n);. Put #include <stdio.h> at the top.

Show example solution

#include <stdio.h>

int main(void) {
    int scores[5] = {85, 92, 78, 90, 88};
    int n = sizeof(scores) / sizeof(scores[0]);
    printf("%d\n", n);
    return 0;
}

Test yourself

Match the sizeof pieces

sizeof(scores)

Total bytes
sizeof(scores[0])

Single element bytes
sizeof(int)

Type bytes
/ sizeof

Division
How many bytes is sizeof(scores) for int scores[5]?
```
int scores[5];
sizeof(scores);
```
- 4
- 5
- 10
- 20
Show answer

20 — 5 int × 4 bytes = 20.

Complete the length idiom

int n = sizeof(scores) / sizeof(____);

Show answer

int n = sizeof(scores) / sizeof(scores[0]);

Next lesson: Multidimensional arrays

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