Question

#include<stdio.h>

int main (void)
{
  int i=257;
  int *ptr=&i;

  printf("%d%d",*((char*)ptr),*((char*)ptr+1));
  return 0;
}

Will the output of the above code implementation defined and the output will vary between little endian and big endian machine?

Answer 1

Yes, it will be implementation defined behaviour.

When you cast ptr with (char *) and then access the first memory location, the first byte stored in memory will be accessed, which will depend on if system is a big-endian or a little-endian system.

Before calling the printf function the arguments are evaluated. And this evaluation as said above will get different values for different systems. Therefore it is implementation defined.