Consider the following code:
#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;
}
What will be the output of the above code?

Correct option is A

The f1 function attempts to swap values locally within its scope, having no effect on a and b. However, f2 swaps the pointers by dereferencing them, effectively changing the values of a and b in main().
· First printf: f1 has no effect, so output remains as ONE TWO.
· Second printf: f2 swaps the values to TWO ONE.
Information Booster:
· Pointer Functions: Demonstrates pass-by-value vs. pointer dereferencing.
· Local Swap Limitation: f1 only affects local copies, not original pointers.