Byte And Bit Ordering

Question

My machine uses little endian byte ordering which means the least significant byte is stored in the lowest address, But what about the bit ordering, will it also be in little endian format?

For example with Ascii encoding where i store A and B, then A will be stored in the lowest address while b will be stored in the highest address.

A = 0x41

B = 0x42

Then what about the bit ordering because I use little endian, does that also mean 0x41 and 0x42 will be stored in little endian bit ordering format?

Answer 1

Yes, indeed, there is both a bit- and a byte-ordering. In most cases, neither of them matters.

Byte ordering only matters if you look at a piece of memory in two different ways. Assume int32 and int8 are 4-byte and 1-byte integers on my computer

int32 x = 0x12345678;
int8 y = *(int8 *)&x;

You stored 0x12345678 into a 4-byte location, and then want to know the contents of one of those bytes. Endian-ness matters. Unless you are writing code that is attempts to serialize a data structure into bytes, you will probably never need to do this. In 99% of the code I've written, the low-level endianness was irrelevant.

Likewise, the X86 instruction has an instruction that says "extract bits x to y", where x and y are hardwired into the instructions. For this case, it is important to know how the hardware orders bits. But unless you are writing in assembly language, you'll never need to know the bit ordering of your machine, and you'll probably write shift-and-mask code to extract bits, and hope the compiler is really good.

So yes, there is bit ordering. But to most programmers, it just doesn't matter.