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0 pointsapi24d ago0 comments

ARM is usually bi-endian, and almost always run in little endian mode. All Apple ARM is LE. Not sure about Android but I’d guess it’s the same. I don’t think I’ve ever seen BE ARM in the wild.

Big endian is as far as I know extinct for larger mainstream CPUs. Power still exists but is on life support. MIPS and Sparc are dead. M68k is dead.

X86 has always been LE. RISC-V is LE.

It’s not an arbitrary choice. Little endian is superior because you can cast between integer types without pointer arithmetic and because manually implemented math ops are faster on account of being linear in memory. It’s counter intuitive but everything is faster and simpler.

Network data and most serialization formats are big endian by convention, a legacy from the early net growing on chips like Sparc and M68k. If it were redone now everything would be LE everywhere.

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amluto23d ago

> Little endian is superior because you can cast between integer types without pointer arithmetic

I’ve heard this one several times and it never really made sense. Is the argument that y you can do:

    short s;
    long *p = (long*)&s;

Or vice versa and it kind of works under some circumstances?

elehack23d ago

Yes. In little-endian, the difference between short and long at a specific address is how many bytes you read from that address. In big-endian, to cast a long to a short, you have to jump forward 6 bytes to get to the 2 least-significant bytes.

lxgr23d ago

Wow, I've been living life assuming that little endian was just the VHS of byte orders with no redeeming qualities whatsoever until today. This actually makes sense, thank you!

hyc_symas23d ago

Network data and most serialization formats are big endian because it's easiest to shift bits in and out of a shift register onto a serial comm channel in that order. If you used little endian, the shifter on output would have to operate in reverse direction relative to the shifter on input, which just causes stupid inconsistency headaches.

jcelerier23d ago

Isn't the issue with shift registers related to endianness at the bit level, while the discourse above is about endianness at the byte level? Both are pretty much entirely separate problems

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