Setting individual indices of a char array equal to ints - c

Question

#include <stdio.h>
int any(char s1[], char s2[]) {
//converts to lower case
int c = 'a';
int i1 = 0;
while (s1[i1] != '\0') {
if (s1[i1] >= 'A' && s1[i1] <= 'Z')
s1[i1] += 32;
++i1;
}
int i2 = 0;
while (s2[i2] != '\0') {
if (s2[i2] >= 'A' && s2[i2] <= 'Z')
s2[i2] += 32;
++i2;
}
i1 = 0;
while (s1[i1] != '\0') {
i2 = 0;
while (s2[i2] != '\0') {
if (s1[i1] == s2[i2])
return i1;
++i2;
}
++i1;
}
return -1;
}
main() {
//printf("test");
printf("%d", any("This is fun", "fin"));
}
This code causes a segmentation fault and I'm pretty sure it happens when I try to set one of the chars in the array equal to an int. How do I not get a seg fault?

Answer 1

You are calling any with pointers to string constants. Attempting to modify these strings invokes undefined behavior.
Also note that the prototype for main should be int main(void) or int main(int argc, char *argv[]), and main should return 0 for successful operation.
You are implementing a case insensitive version of a generalized version of strchr with multiple characters, but you should not modify the argument strings, and you should rely on the functions from <ctype.h> instead of assuming ASCII.
Here is a better version:
#include <ctype.h>
#include <stdio.h>
int any(const char s1[], const char s2[]) {
int i1 = 0;
while (s1[i1] != '\0') {
int i2 = 0;
while (s2[i2] != '\0') {
if (tolower((unsigned char)s1[i1]) == tolower((unsigned char)s2[i2]))
return i1;
++i2;
}
++i1;
}
return -1;
}
int main(void) {
//printf("test");
printf("%d", any("This is fun", "fin"));
return 0;
}