C code for counting number of occurrences of a word in string

Question

I am trying to write a function that accepts a string and a word, and output the number of times it appears inside the string.

My bid:

int CountWord(char *s, char *word) {
    char first = word[0];
    int i, j, count = 0;
    while (s[i] != '\0') {
        if (s[i] == first)
            for (j = 1; (s[i+j] == word[j]) && (word[j] != '\0') && (s[i+j] != '\0'); ++j) {
                if (word[j] == '\0')
                    ++count;
        }
    }
    return count;
}

The problem: It doesn't work on "code-blocks".

Thanks in advance for your appreciated help.

Answer 1

There are multiple problems in your code:

• you do not initialize i
• you do not increment i
• the test (word[j] != '\\0') in the for loop prevents the counting code from ever executing.
• a match cannot be found at the end of the string
• the count will be incorrect if word is the empty string.
• int might not have the required range for very long strings on some architectures.

Here is a corrected version:

size_t CountWord(const char *s, const char *word) {
    char first = word[0];
    size_t i, j, count = 0;

    if (first == '\0')
        return strlen(s) + 1;

    for (i = 0; s[i] != '\0'; i++) {
        if (s[i] == first) {
            for (j = 1; word[j] != '\0' && s[i + j] == word[j]; j++)
                continue;
            if (word[j] == '\0')
                count++;
        }
    }
    return count;
}

Note that this implementation will return 2 for CountWord("aaa", "aa") which may or may not be the expected result. The specification must be precise and tell whether to count overlapping occurrences of word or just non overlapping matches.