#include<stdio.h>
void f1(char *x, char *y) {
char *t1;
t1 = x;
x = y;
y = t1;
}
void f2(char **x, char **y) {
char *t1;
t1 = *x;
*x = *y;
*y = t1;
}
int main() {
char *a = "ONE", *b = "TWO";
f1(a, b);
printf("%s %s ", a, b);
f2(&a, &b);
printf("%s %s", a, b);
return 0;
}

1. AONE TWO TWO ONE
2. BTWO ONE ONE TWO
3. CONE TWO ONE TWO
4. DTWO ONE TWO ONE

Pick an option to check yourself, or open the answer below.

Show answer & explanation

Correct answer: A. ONE TWO TWO ONE

Explanation

f1 swaps only its local copies of the pointers, so main still prints ONE TWO. f2 receives the addresses of a and b and swaps the pointers themselves, so the second print is TWO ONE.