Question: I haven't found a specific question here on SO, if this is a duplicate please point it out to me and I'll delete this.

So really, does endianness have anything to do with **bit** order?

[This](http://www.umsl.edu/~ewm4g2/articles/endian_swap.html) seems to imply the answer is NO, while other sources (I fail to find one right now, but surely I've read some time ago some articles) imply that endianness is the order of **bytes and bits**.

To be more specific: in a Big Endian architecture, where MSB is first, inside any byte, is also MSb first? Conversly, on Little Endian systems, where LSB is first, is LSb also first?

LAST EDIT: I found [this](http://www.linuxjournal.com/article/6788) which says "**Bit order usually follows the same endianness as the byte order for a given computer system**"

Answer: 
Endianness applied to only byte order. Not for bit order. The bit order remains same.

Why?

Memory is byte-addresseable. This is just a fancy way of saying that each address stores one byte. So You can change the order of bytes, not bits in memory.

Endianness only makes sense when you want to break a large value (such as a word) into several small ones. You must decide on an order to place it in memory.

Endianness only makes sense when you are breaking up a multi-byte quantity, and attempting to store the bytes at consecutive memory locations. But if you take in a register, it doesn't make sense. A register is simply a 32 bit quantity(depends on your processor/controller) and endianness does not apply to it.

Bit order usually follows the same endianness as the byte order for a given computer system - It is True

For more info [Endianness](http://www.cs.umd.edu/class/sum2003/cmsc311/Notes/Data/endian.html)
Endianness works on the basis of bytes not bits. Bits are not addressable.