C strings as char arrays, explained

Strings are char arrays

C has no real string type. A "string" is just an array of char with one rule: it ends with a zero byte, written '\0'. That terminator is how every string function knows where the text stops.

char name[] = "Ada";

This creates a 4-byte array, not 3:

index:  [0]   [1]   [2]   [3]
char:   'A'   'd'   'a'   '\0'
byte:    65   100    97     0

Yes, characters are numbers. 'A' is just 65 in ASCII, and %c prints the character while %d prints its number. That means you can do arithmetic on letters: 'a' - 'A' is 32, the gap between lowercase and uppercase.

The standard function strlen (from <string.h>) counts the characters before the '\0'. There is no stored length anywhere; it really walks the array until it hits zero. So strlen("Ada") is 3 while sizeof(name) is 4.

This design is simple and cheap, and also the source of decades of bugs. If the '\0' is missing, functions like strlen and printf("%s") keep reading past the end of the array into whatever memory follows: undefined behaviour again. Always leave room for the terminator when you size a char array yourself.

One more trap: char *s = "hi"; points at a string literal, which you must not modify (undefined behaviour). char s[] = "hi"; makes your own copy in an array, which you can change freely.

Try it yourself

Edit it. Break it. Run it again.

#include <stdio.h>
#include <string.h>

int main(void) {
    char word[] = "Hi!";
    printf("strlen says %zu, sizeof says %zu\n", strlen(word), sizeof(word));
    for (int i = 0; i < (int)sizeof(word); i++) {
        printf("word[%d] = %d", i, word[i]);
        if (word[i] != '\0') {
            printf(" ('%c')", word[i]);
        }
        printf("\n");
    }
    return 0;
}

Your turn

Type it yourself. That is the whole trick.

1strlen by hand

Finish int my_strlen(char s[]) so it counts characters until it reaches '\0', without using <string.h>. The program should print:

5
0
13

#include <stdio.h>

int my_strlen(char s[]) {
    int n = 0;
    return n;
}

int main(void) {
    printf("%d\n", my_strlen("hello"));
    printf("%d\n", my_strlen(""));
    printf("%d\n", my_strlen("pointers soon"));
    return 0;
}

2SHOUT

Change every lowercase letter in msg to uppercase, in place, using arithmetic on chars (no toupper). Leave other characters alone. Then print it: STOP COPYING, START TYPING!

#include <stdio.h>

int main(void) {
    char msg[] = "stop copying, start typing!";

    printf("%s\n", msg);
    return 0;
}