Question

at the moment, I’m trying some pointer stuff in C. But now, I have a problem with a pointer array. By using my code below, I get a strange output. I think there is a big mistake in the code, but I can’t find it.

I just want to print the strings of the pointer array.

#include <stdio.h>


int main(void)
{
    char *words[] = {"word1", "word2", "word3"};
    char *ptr;
    int i = 0;

    ptr = words[0];

    while(*ptr != '\0')
    {
        printf("%s", *(words+i));
        ptr++;
        i++;
    }

    return 0;
}

Output: word1word2word3Hã}¯Hɡ

Thanks for helping.

Answer 1

while(*ptr != '\0')
{
    printf("%s", *(words+i));
    ptr++;
    i++;
}

initially, ptr points to the 'w' in "word1". So the loop iterates five times until *ptr == '\0'. But the array words contains only three elements, thus the fourth and fifth iteration invoke undefined behaviour and garbage is printed when the bytes after the words array are interpreted as pointers to 0-terminated strings. It could easily crash, and if you try it on other systems, with other compilers or compiler settings, it will sometimes crash.

You could translate the loop to

for(i = 0; i < strlen(words[0]); ++i) {
    printf("%s", words[i]);
}

to see more easily what it does.

If you want to print out the strings in the words array, you can use

// this only worls because words is an actual array, not a pointer
int numElems = sizeof words / sizeof words[0];
for(i = 0; i < numElems; ++i) {
    printf("%s", words[i]);
}

As words is an actual array, you can obtain the number of elements it contains using sizeof. Then you loop as many times as the array has elements.