What will be the output of the program?

#include<stdio.h>
void swap(char *, char *);

int main()
{
    char *pstr[2] = {"Hello", "CuriousTab"};
    swap(pstr[0], pstr[1]);
    printf("%s\n%s", pstr[0], pstr[1]);
    return 0;
}
void swap(char *t1, char *t2)
{
    char *t;
    t=t1;
    t1=t2;
    t2=t;
}

Step 2: char *pstr[2] = {"Hello", "CuriousTab"}; The variable pstr is declared as an pointer to the array of strings. It is initialized to

pstr[0] = "Hello", pstr[1] = "CuriousTab"

Step 3: swap(pstr[0], pstr[1]); The swap function is called by "call by value". Hence it does not affect the output of the program.

If the swap function is "called by reference" it will affect the variable pstr.

Step 4: printf("%s\n%s", pstr[0], pstr[1]); It prints the value of pstr[0] and pstr[1].

Hence the output of the program is

Hello
CuriousTab