OneAndOneIs2

K&R exercise 2.5 - create a function that finds the first occurence of a character in string 1 that is also present in string 2, and returns either the position (if a match is found) or -1 (if not)

I wrote it, and tested it, but it never returned the -1 result. Because, of course, when using getline() to get the two strings, there's always a carriage return at the end of each string. Took me a few moments to realize this, and then I breathed a sigh of relief that I hadn't made a mistake after all.

The exercise doesn't mention "except for '\n's" so I left it as-is, but I did test that it works properly if you crop out the '\n', just to be sure.

Aside from that slight oddity, it was a remarkably simple exercise, and most of the code was recycled from the previous example. . .

Follow up:

/* K&R exercise 2.5: Write any(s1, s2) that returns the first location in s1
 * where any character in s2 occurs, or -1 if no match is found.
 * The exercise makes no mention of excluding '\n' so all lines of sub-MAXLINE
 * length will return a valid result in this program */
#include <stdio.h>

#define MAXLINE 500

int getline(char line[], int maxline);
signed int any(char s1[], char s2[]);

int main()
{
    char string1[MAXLINE];
    char string2[MAXLINE];
    int i;

    printf("Give me a string:\n");
    getline(string1, MAXLINE);
    printf("Give me the comparison string:\n");
    getline(string2, MAXLINE);
    i = any(string1, string2);
    if (i == -1)
        printf("No match found\n");
    else
        printf("The character at array position %d, \'%c\', is present in both strings\n", i, string1[i]);
    return 0;
}

int getline(char line[], int maxline)
{
    int i, c;

    for (i=c=0; c != '\n' && (c = getchar()) != EOF && i < maxline-1; ++i)
        line[i] = c;
    line[i] ='\0';
    return i;
}

signed int any(char s1[], char s2[])
{
    int i, j;
    /* Match should default to -1 */
    signed int match = -1;

    /* Loop through the characters in s1 */
    for (i = 0; s1[i] != '\0' && match == -1; ++i)
    {
        j = 0;
        /* Check each character in s1 against all charcters in s2 */
        while (s2[j] != '\0' && match == -1)
        {
            if (s1[i] == s2[j])
                /* If a match is found, set match==i */
                match = i;
            ++j;
        }
    }
    return match;
}