Exercise 5.4 - strend returns 1 if string t occurs at the end of the string s¶

Question¶

Write the function strend(s,t), which returns 1 if the string t occurs at the end of the string s, and zero otherwise.

#include<stdio.h>

#define MAXLINE 1000

int mgetline(char s[], int lim);

int strend(char *s, char *t);

int mystrlen(char *t);

int main(void) {
    char s[MAXLINE], t[MAXLINE];
    int ret;
    mgetline(s, MAXLINE);
    mgetline(t, MAXLINE);
    ret = strend(s, t);
    printf("%d", ret);
    return 0;
}

int mgetline(char s[], int lim) {
    int c, i;

    for (i = 0; i < lim - 1 && ((c = getchar()) != EOF) && c != '\n'; ++i)
        s[i] = c;

    if (c == '\n') {
        s[i] = c;
        ++i;
    }
    s[i] = '\0';

    return i;
}

int strend(char *s, char *t) {
    int len;
    len = mystrlen(t);
    while (*s != '\0')
        ++s;
    --s;

    while (*t != '\0')
        ++t;

    --t;
    while (len > 0) {
        if (*t == *s) {
            --t;
            --s;
            --len;
        } else
            return 0;
    }
    if (len == 0)
        return 1;
}

int mystrlen(char *t) {
    char *p;
    p = t;

    while (*p != '\0')
        ++p;

    return p - t;
}

Explanation¶

This program determines if the string t occurs at the end of string s. So the output of the program will look like.

$ ./a.out
something
thing
1

$ ./a.out
something
non
0

The primary part of this program is the strend function, which takes two character pointers, s and t. It calculates the length of t and stores in the variable len. And then, we back off till the last characters in both s and t.

while(*s!='\0')
    ++s;
--s;

while(*t!='\0')
    ++t;

--t;

And then we look for the match from the end. This is checked in this while loop. While the len is > 0, check if s and t are same and back off one character at a time.

while(len > 0)
{
    if(*t==*s)
    {
        --t;
        --s;
        --len;
    }
    else
        return 0;
}
if( len == 0)
    return 1;

If the string t exhausts, that is, it’s length, len becomes 0, then we known that string t occurs at the end of string s and we return 1. Otherwise, we return 0.