output of the program

[geekocoder]

can anyone please explain me the output of the program

#include<stdio.h>
main()
{
char *p="hai friends",*p1;
p1=p;
while(*p!='\0') ++*p++;
printf("%s %s",p,p1);
getchar();
}

[ne555]

"whatever" is a const char*

[vlad from moscow]

I think you cannot understand expression ++*p++;
To understand it I will rewrite it as a sequence of actions

p++; // returns current value of the pointer and increases it

*p++; // dereferences the pointer so it correcponds to a value of type char

++*p++; increases the value in the object of type char.

For example if *p++ is equal to 'h' (the first letter of the character literal) then ++*p++ will increase 'h' and its value will become 'i'

So the loop while(*p!='\0') ++*p++; tries to change the string literal "hai friends" to something as "ibj gsjfoet"

However you may not change string literals in C/C++. So the behaviour of the program is undefined and can result in throwing an exception.

Moreover if to accept that the program will not throw an exception pointer p will points to '\0' after the loop. So in the next statement printf("%s %s",p,p1); there will be displaying nothing for the first argument.

The correct code shall look the following way. I am assuming that it is written in C

#include<stdio.h>

int main( void )
{
   char s[] = "hai friends";
   char *p = s;
   
   printf( "%s\n",  s );

   while (*p != '\0' ) ++*p++;

   printf( "%s\n",  s );

   getchar();
}

EDIT: I removed some typo from the code.:)

[geekocoder]

why is the expression been parsed from right to left...why isnt precedence working here.....p should be dereferenced 1st and the postfix operation and then prefix???

[vlad from moscow]

The postfix operator has the highest priority among others operators in the expression. All other operators in the expression are parsed from right to left. There are such rules in C/C++.