OneAndOneIs2

They give you the function "strindex" that basically scans a line for a pattern, grep-like, and returns the location of the first occurrence of the pattern.

Then they tell you to write "strrindex" to return the position of the RIGHTMOST occurrence of the pattern - i.e. if it occurs twice, they want the position of the second occurrence.

I was going to take their code and modify it by adding variables to make a note of each time a pattern was found. But then I thought "Why not just start at the right and work leftwards? Then I only have to look for the first pattern."

So I sat down with pen & paper and worked out the logic. And then it seemed silly NOT to go ahead and write it to test it worked.

Took a few more scribblings to work out the exact logic because of the numbering that applies to strings: "abc" is a four-character string because of the '\0' at the end, but the position numbers of the four-character string only go up to 3 because numbering starts at zero, and so confusion can happen..

But once I got that sorted, and I'd slapped myself for using [i] instead of [i++] with the result that I had an infinite loop, the code worked flawlessly. To my surprise, it must be said.

[More:]

int strrindex(char s[], char t[])
{
        int i, j, k, slength, tlength;

        slength = 0;
        tlength = 0;

        /* Find the length of the s and t strings */
        i = 0;
        while (t[i++] != '\0')
                ++tlength;

        i = 0;
        while (s[i++] != '\0')
                ++slength;

        k = 1;
        /* Start from the right and work left */
        for (i = slength; k > 0 && i > 0; i--)
                /* Compare t to s until we have a match */
                for (j=(i-1), k=tlength; k > 0 && s[j]==t[k-1]; j--,k--)
                        ;
        if (k == 0)
                return (j+1);
        else
                return -1;
}