Exercise 2.5 - return the first location in the string s1 comparing s2#
Question#
Write the function any(s1,s2), which returns the first location in a string s1 where any character from the string s2 occurs, or -1 if s1 contains no characters from s2. (The standard library function strpbrk does the same job but returns a pointer to the location.)
/**
* Exercise 2.5
*
* Write the function any(s1,s2) which returns the first location in the string
* s1 where any character from the string s2 occurs, or -1 if s1 contains
* no characters from s2. ( The standard library function strpbrk does
* the same job but retuns a pointer to the location
*
**/
#include<stdio.h>
#define MAXLINE 1000
int mgetline(char line[],int maxline);
int any(char s1[],char s2[]);
int main(void)
{
char s1[MAXLINE],s2[MAXLINE];
int val;
/* Give the first string s1 */
mgetline(s1,MAXLINE);
/* Give the second string s2 */
mgetline(s2,MAXLINE);
val = any(s1,s2);
printf("%d",val);
return 0;
}
int mgetline(char s[],int lim)
{
int i,c;
for(i=0;i<lim-1 && (c=getchar())!=EOF && c!='\n';++i)
s[i]=c;
if(c=='\n')
s[i++]=c;
s[i]='\0';
}
int any(char s1[],char s2[])
{
int i,j;
for(i=0;s1[i]!='\0';++i)
{
// iterate through s2 while trying to find matching character from s1
for(j=0;(s1[i]!=s2[j]) && s2[j]!='\0';++j)
; // continue
if(s2[j]!='\0' && s2[j] != '\n') { // check that s2 [j]! = '\n', since s1 and s2 both have the character '\n' in the penultimate position of the string.
return i;
}
}
return -1;
}
Explanation#
The important part of the program is the function any which takes two strings s1 and s2 and tries to find if any character in s2 matches s1. We set a flag, check_next_char which is toggled to 0 if we find the match, otherwise we have it as 1.
The first for loop iterates through all the characters in s1 while the condition check_next_char is 1. In the second for loop, if we find that a char in s2 matches s1, that is s2[j] == s1[i] and s2 has not reached EOL, then we set check_next_char to 0. That is we found a match at i and we return that.
If we dont find a match in s2, we increment i and take the next character from s1. If dont find a match at all, then we return -1.