Is this code Implementation defined?

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. The classic way to detect endianness at run time is the following way:

uint32_t var = 1;
uint8_t *ptr = (uint8_t*)&var;

if(*ptr) puts("Little Endian");
else puts("Big Endian");

Even though 257 => 0x0101, an int is most likely 32 bits in which case on a BE machine it will print 00 and on LE, 11.

Answer 2

Yes it will. I always write it this way since I am never sure about operator precedence:

*(((char*)ptr)+1)

And to achieve what you want change your %d to a %c in the format string.