C · Course navigation

C strings and null terminators

Traverse characters without reading beyond the string.

On this page

Before you start

Arrays, char, and loops.

Traverse characters without reading beyond the string.
Predict the sample output and explain the state changes.
Complete the practice task and check the boundary cases.

The concept

A C string ends with a zero-valued character, written '\0'. A string literal used to initialize a char array supplies that terminator automatically.

The loop counts selected vowels until the terminator. This example uses lowercase ASCII text and counts a, e, and i to keep the condition visible.

Code example

#include <stdio.h>
#include <assert.h>
int main(void) {
    const char text[] = "learning";
    int count = 0;
    for (int i = 0; text[i] != '\0'; i++) {
        if (text[i] == 'a' || text[i] == 'e' || text[i] == 'i') {
            count++;
        }
    }
    printf("%d\n", count);
    assert(count == 3);
    return 0;
}

Output

Example details

Follow the code

Open the walkthrough to inspect the example one step at a time.

Show guided walkthrough

Guided walkthroughReady

1#include <stdio.h>

2#include <assert.h>

3int main(void) {

4 const char text[] = "learning";

5 int count = 0;

6 for (int i = 0; text[i] != '\0'; i++) {

7 if (text[i] == 'a' || text[i] == 'e' || text[i] == 'i') {

8 count++;

9 }

10 }

11 printf("%d\n", count);

12 assert(count == 3);

13 return 0;

14}

Follow the example

Start to inspect the source, values, and output. These prepared steps explain the published sample; grouped loop steps describe multiple iterations.

View the complete dry-run table

Prepared trace of the published program. Some rows group repeated iterations.
What happens	State after step	Output so far
Store eight letters followed by the terminator.	`text = l e a r n i n g \0`	—
The e at index 1 matches.	`count = 1`	—
The a at index 2 matches.	`count = 2`	—
The i at index 5 matches.	`count = 3`	—
Index 8 reaches the terminator and stops the loop.	`count = 3`	3

Behind the scenes

How C processes this code

The C compiler produces a native program. Variables occupy memory, and execution enters main.

Cost and common mistakes

O(n) time and O(1) auxiliary space.

A char array without a terminator is not a valid C string.
Byte-wise character handling does not implement general Unicode text processing.

Try it yourself

Extend the condition to include o and u, and use 'education'.

Reveal answer and reasoning

The count is 5: e, u, a, i, o.