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?
If it is little-endian, it would be stored as “01 00 00 00”. The program checks the first byte by dereferencing the cptr pointer. If it equals to 0, it means the processor is big-endian(“00 00 00 01”), If it equals to 1, it means the processor is little-endian (“01 00 00 00”).
Intel based processors are little endians. ARM processors were little endians. Current generation ARM processors are bi-endian.
Solely big-endian architectures include the IBM z/Architecture and OpenRISC. Some instruction set architectures are "bi-endian" and allow running software of either endianness; these include Power ISA, SPARC, ARM AArch64, C-Sky, and RISC-V.
All versions of Windows that you'll see are little-endian, yes. The NT kernel actually runs on a big-endian architecture even today.
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; }
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