c++ - char s1[ ] = " xyz" and char *s1 = "xyz"

Question

I am trying to initialize s1 with the value and later in squeeze() . I am changing s1. Pointer initialization throws a seg fault but array version works fine. Could someone tell me If i can do it with the pointer version and what is wrong with this ? may be because i have not used malloc ??

int main() {

        char s1[] = "xyz abc zzz" ; // this works fine.
      // char *s1 = "xyz abc zzz"; // if i initialize like this it throws a seg fault. 

         char *s2 = "abx xxx xxx" ;
        squeeze(s1,s2) ;
        puts (s1) ;
        return 0 ;

}

void squeeze(char *s1, char *s2 ) {

        int i , j, k = 0;

        for (i=0 ; s1[i] != '\0' ; ++i) {

                for (j=0; s2[j] != '\0'; j++) {
                        if (s2[j] == s1[i])
                                break;


                }
           if(s2[j] == '\0')

                s1[k++] = s1[i] ;

        }

        s1[k++] = '\0' ;

}

Answer 1

The difference is:

Declaration:

char s1[] = "xyz abc zzz" ; 

Creates an array of size strlen(RHS) + 1 and is modifiable. In second form

char *s1 = "xyz abc zzz"; 

s1 is a pointer to a string literal which is in an unmodifiable memory area; hence modification of the string is illegal operation.