Use pointer math instead of array indexing

Question

I'm trying to solve a problem found on my C programming book.

#include <stdio.h>
char *f (char s[], char t[], char out[]);
int main(void)
{
    char s[] = "ISBN-978-8884981431";
    char t[] = "ISBN-978-8863720181";
    char out[10];
    printf ("%s\n", f(s,t,out));
    return 0;
}
char *f (char s[], char t[], char out[]) {
    int i;
    for (i=0; s[i] == t[i]; i++)
    out[i] = s[i];
out[i] = '\0';
return &out[0];
}

As you can see from the code, this code uses as return value &out[0]: does this means that the complete array is used as return value?

char *f (char s[], char t[], char out[]);
int main(void)
{
    char s[] = "ISBN-978-8884981431";
    char t[] = "ISBN-978-8863720181";
    char out[10];
    printf ("%s\n", f(s,t,out));
    return 0;
}
char *f (char s[], char t[], char out[]) {
    for (; *(s+=1) == *(t+=1);)
            *(out+=1) = *s;
    *(out+1) = '\0';
    return out;
}

This is my proposed solution, but while the proposed code returns "ISBN-978-88", mine only returns "8". The array is smaller than the lenght of the string, how the proposed code can work without any kind of overflow? Thanks for your responses.

Answer 1

Your code is too aggressive on side effects: the += 1 operation (which is more commonly denoted simply as ++) should be applied after the copy to the output has been made, not after the comparison.

In addition, you need to save the value of the out buffer before incrementing the pointer, so that you could return a pointer to the beginning of the copied string.

char *orig = out;
for ( ; *s == *t ; s++, t++)
    *out++ = *s;
*out = '\0';
return orig;

Demo on ideone.