Bit order in C/C++

Question

I have to implement a protocol which defines data in 8bit words, which starts with the least significant bit (LSB) first. I want to realize this data with unsigned char, but I don't know what's the bit order of LSB and most significant bit (MSB) in C/C++, that could possible require swapping the bits.

Can anybody explain me how to find out an unsigned char is encoded: with MSB-LSB or LSB-MSB?

Example:

unsigned char b = 1;

MSB-LSB: 0000 0001 LSB-MSB: 1000 0000

Answer 1

Endian-ness is platform dependent. Anyway, you don't have to worry about actual bit order unless you are serializing the bytes, which you may be trying to do. In which case, you still don't need to worry about how individual bytes are stored while they're on the machine, since you will have to dig the bits out individually anyway. Fortunately, if you bitwise AND with 1, you get the LSB, regardless of storage order; bit-AND with 2 and you get the next most significant bit, and so on. The compiler will sort out what constants to generate in the machine code, so that level of detail is abstracted away.

Answer 2

There is no such thing in C/C++. The least significant bit is -- well -- the least significant bit. Since the bits don't have addresses, there is no other ordering.