Question

#include <stdio.h>

int main()
{
 char *arr = "This is to test";
 printf("\n%c %c ",*(arr++), *(arr++));
 return 0;
}

This program outputs h T rather than outputting h h. Why is it so?

Answer 1

arr++ is equivalent to arr += 1. You are actually incrementing arr. This is what is happening in your situation.

printf("\n%c %c ",*(arr++), *(arr++));
//                  ^         ^Evaluates to 0 therefore prints T then increments by 1
//                  ^Evaluates to 1 therefore prints h and then increments by 1
//Now if where to prinf %s arr it should print "is is to test"

What you want to do is

printf("\n%c %c ",*(arr+1), *(arr+1));

Note: As others have pointed out multiple pre/postfix ++/– produce undefined behavior (order of evaluation) and should be avoided, although I understand in your situation you were not trying to modify the original array.