CODEWITHSUNDEEP
cx86intelendiannessamd-processor
Frank V
Frank V

Reputation: 25429

Does my AMD-based machine use little endian or big endian?

I'm going though a computers system course and I'm trying to establish, for sure, if my AMD based computer is a little-endian machine? I believe it is because it would be Intel-compatible.

Specifically, my processor is an AMD 64 Athlon x2.

I understand that this can matter in C programming. I'm writing C programs and a method I'm using would be affected by this. I'm trying to figure out if I'd get the same results if I ran the program on an Intel based machine (assuming that is little endian machine).

Finally, let me ask this: Would any and all machines capable of running Windows (XP, Vista, 2000, Server 2003, etc) and, say, Ubuntu Linux desktop be little endian?

Upvotes: 37

Views: 41908

Answers (9)

sleep
sleep

Reputation: 4954

We now have std::endian!

constexpr bool is_little = std::endian::native == std::endian::little;

https://en.cppreference.com/w/cpp/types/endian

Upvotes: -1

tf3
tf3

Reputation: 467

The below snippet of code works:

#include <stdio.h>

int is_little_endian() {
  short x = 0x0100; //256
  char *p = (char*) &x;
  if (p[0] == 0) {
    return 1;
  }
  return 0;
}

int main() {
  if (is_little_endian()) {
    printf("Little endian machine\n");
  } else printf("Big endian machine\n");
  return 0;
}

The "short" integer in the code is 0x0100 (256 in decimal) and is 2 bytes long. The least significant byte is 00, and the most significant is 01. Little endian ordering puts the least significant byte in the address of the variable. So it just checks whether the value of the byte at the address pointed by the variable's pointer is 0 or not. If it is 0, it is little endian byte ordering, otherwise it's big endian.

Upvotes: 4

Mehrdad Afshari
Mehrdad Afshari

Reputation: 422280

All x86 and x86-64 machines (which is just an extension to x86) are little-endian.

You can confirm it with something like this:

#include <stdio.h>
int main() {
    int a = 0x12345678;
    unsigned char *c = (unsigned char*)(&a);
    if (*c == 0x78) {
       printf("little-endian\n");
    } else {
       printf("big-endian\n");
    }
    return 0;
}

Upvotes: 80

Linas
Linas

Reputation: 11

/* by Linas Samusas  */

#ifndef _bitorder 
#define _bitorder 0x0008

#if (_bitorder > 8)
#define BE
#else
#define LE
#endif

and use this

#ifdef LE
#define Function_Convert_to_be_16(value)  real_function_to_be_16(value)
#define Function_Convert_to_be_32(value)  real_function_to_be_32(value)
#define Function_Convert_to_be_64(value)  real_function_to_be_64(value)
#else
#define Function_Convert_to_be_16
#define Function_Convert_to_be_32
#define Function_Convert_to_be_64
#endif

if LE 

unsigned long number1 = Function_Convert_to_be_16(number2);

*macro will call real function and it will convert to BE

if BE 

unsigned long number1 = Function_Convert_to_be_16(number2);

*macro will be defined as word not a function and your number will be between brackets

Upvotes: 1

Ben Hoyt
Ben Hoyt

Reputation: 11044

Assuming you have Python installed, you can run this one-liner, which will print "little" on little-endian machines and "big" on big-endian ones:

python -c "import struct; print 'little' if ord(struct.pack('L', 1)[0]) else 'big'"

Upvotes: 12

ephemient
ephemient

Reputation: 205014

"Intel-compatible" isn't very precise.

Intel used to make big-endian processors, notably the StrongARM and XScale. These do not use the IA32 ISA, commonly known as x86.

Further back in history, Intel also made the little-endian i860 and i960, which are also not x86-compatible.

Further back in history, the prececessors of the x86 (8080, 8008, etc.) are not x86-compatible either. Being 8-bit processors, endianness doesn't really matter...

Nowadays, Intel still makes the Itanium (IA64), which is bi-endian: normal operation is big-endian, but the processor can also run in little-endian mode. It does happen to be able to run x86 code in little-endian mode, but the native ISA is not IA32.

To my knowledge, all of AMD's processors have been x86-compatible, with some extensions like x86_64, and thus are necessarily little-endian.

Ubuntu is available for x86 (little-endian) and x86_64 (little-endian), with less complete ports for ia64 (big-endian), ARM(el) (little-endian), PA-RISC (big-endian, though the processor supports both), PowerPC (big-endian), and SPARC (big-endian). I don't believe there is an ARM(eb) (big-endian) port.

Upvotes: 10

FCo
FCo

Reputation: 457

An easy way to know the endiannes is listed in the article Writing endian-independent code in C

const int i = 1;
#define is_bigendian() ( (*(char*)&i) == 0 )

Upvotes: 16

mnuzzo
mnuzzo

Reputation: 3577

You have to download a version of Ubuntu designed for big endian machines. I know only of the PowerPC versions. I'm sure you can find some place which has a more generic big-endian implementation.

Upvotes: 2

1800 INFORMATION
1800 INFORMATION

Reputation: 135463

In answer to your final question, the answer is no. Linux is capable of running on big endian machines like e.g., the older generation PowerMacs.

Upvotes: 3

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