Thread: Big Endian & Little Endian

Hi Guys,
How to find out the given processor is a big endian or small endian.??

Try this code to find out your processor is big or small endian

Code:

main()
{
   int num = 1;
   if(*(char *)&num == 1)
   {
      printf("\nLittle-Endian\n");
   }
   else
  {
      printf("Big-Endian\n");
  } 
}

Little Endian means that the lower order byte of the number is stored in memory at the lowest address, and the higher order byte is stored at the highest address. That is, the little end comes first.

For example, a 4 byte, 32-bit integer


Byte3 Byte2 Byte1 Byte0


will be arranged in memory as follows:


Base_Address+0 Byte0
Base_Address+1 Byte1
Base_Address+2 Byte2
Base_Address+3 Byte3


Example :Intel processors use "Little Endian" byte order.


"Big Endian" means that the higher order byte of the number is stored in memory at the lowest address, and the lower order byte at the highest address. The big end comes first.


Base_Address+0 Byte3
Base_Address+1 Byte2
Base_Address+2 Byte1
Base_Address+3 Byte0


Motorola, Solaris processors use "Big Endian" byte order.

Hi,
I was searching for the same problem, and this thread proved to be very helpful. Thanks to everyone.
I have just one more query, how can I find whether the computer is little endian or big endian using a union.

I fully understand the code provided by Swaugh (thanks for it) but I also need to find out how I can accomplish the same using a union.

Thanks in anticipation,
regards,
koodoo

Originally Posted by dwks

You shouldn't bump old threads, especially a thread as old as this one. You can start a new thread and link to an old thread if you want to.

I'll remember that, the next I have such a query.

Originally Posted by dwks

BTW, it's [color=red]text[/color], not [red]text[/red].

Thanks for that too, actually I'm used to some other forums where that kinda thing works

Originally Posted by dwks

This might work, though I'm not too sure about it:

Code:

union {
    short snum;
    unsigned char cnum;
} test;

test.snum = 1;
if(test.cnum == 1) puts("little edian");
else puts("bug edian");

I think it will thanks for the help. I must also apologize coz I posted so very early (without putting in much effort myself, I was kinda in a hurry).
I searched and found a similar solution.
The following code :

Code:

#include <stdio.h>

void main()
{
        union s
       {
            int i;
            char ch[2];
       };


union s obj;
obj.i=256;
printf("%d %d %d\n",obj.i,obj.ch[0],obj.ch[1]);

}

would print 256 0 1 on a lil endian machine, while 256 1 0 on a big endian machine.

Thanks for all the help

yeah, thanks again.
I don't know why I am in such a hurry and messing up things.
In the discussion from where I got this code, the first error pointed out was that of using void main() which of course is a bad practise, I forgot to correct it while copying and pasting code.
Moreover, the link posted earlier in this thread http://c-faq.com/misc/endiantest.html
describes both the methods (using pointers & unions) It deals with varied sized ints, however it should have chosen unsigned char as you have suggested.

P.S. On more query, this might be a bit silly, but I could not find how to add a signature to my profile. I tried the USER CP and edit profile links, but it was not there...I feel completely stupid as of now.