Step 6 of 10

Strings through pointers

Strings are char arrays, and a string passed to a function arrives as a char * pointing at its first character. Walking a string with a pointer is the most common pointer code in C, so it's worth getting fluent with it.

#include <stdio.h>

int count_vowels(const char *s) {
    int count = 0;
    for (const char *c = s; *c != '\0'; c++) {
        if (*c == 'a' || *c == 'e' || *c == 'i' || *c == 'o' || *c == 'u') {
            count++;
        }
    }
    return count;
}

int main(void) {
    printf("%d\n", count_vowels("programming in c"));
    return 0;
}

4

Reading the loop

• const char *c = s starts a pointer at the first character.
• *c != '\0' keeps going until the terminator.
• c++ moves to the next character.
• Inside, *c is the current character.

It does the same thing as for (int i = 0; s[i] != '\0'; i++) with s[i]. Both styles are common; the pointer version is idiomatic C and shows up in the standard library's own code.

String literals are read-only

A literal like "programming in c" must never be modified: writing to it is undefined behavior. On most desktop systems literals live in read-only memory, so an attempt crashes; on others it silently misbehaves. You can point at a literal with a const char *, and the const makes the compiler stop you from writing through it. Declaring parameters as const char * lets callers pass literals safely, and the compiler will stop you if the function tries to modify them.

Your turn: write int count_char(const char *s, char ch) that returns how many times ch appears in s.

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