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#include <stdio.h>
/* The C Programming Language: 2nd Edition
*
* Exercise 2-4: Write an alternate version of squeeze(s1, s2) that deletes
* each character in s1 that matches any character in the string s2.
*
* Answer: This one is fairly easy, considering the fun trick that's covered
* in the passage before this exercise with the unary operators ++ and --. C's
* interesting behavior with these operators allows the programmer to write
* shorter, faster logic.
*
* That said, I couldn't find a way to iterate through s2[] and s1[] in a
* single loop, and I needed to use a flag. There may be a more clever way to
* solve this.
*
*/
/* It'd be better to make this return a pointer (or string), but the book */
/* hasn't covered it yet! */
void squeeze(char s1[], char s2[]) {
int i, j, k, match;
for (i = j = 0; s1[i] != '\0'; i++) {
/* I don't see a way to do this without a flag */
match = 0;
for (k = 0; s2[k] != '\0'; k++) {
if (s1[i] == s2[k]) {
match = 1;
break;
}
}
/* check our flag. if there wasn't a match, j needs to match i's value */
if (!match) {
s1[j++] = s1[i];
}
}
s1[j] = '\0';
}
int main() {
char foo[16] = "foobarbaz";
char bar[16] = "boz";
squeeze(foo, bar);
printf("%s\n", foo); // Should read "fara"
return 0;
}
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