Bits and Bytes – Is 8 Bit and 1 Byte a Valid Question?

It has to do with what the CPU can easily address. For example on an x86 processor there is an eax (32 bits), ax (16 bits) and a ah (8 bits) but no single bit register. So in order for it to use a single bit the CPU will have to do a read/modify/write to change the value. If it is stored as a byte a single read or write can be used to inspect/change the value.

Additionally one might wonder if it would be better to use a single bit vs a full byte, after all a byte will be wasting 7 bits. Unless space is a constraint the one should go for the byte because, at least the x86 and I think others, there is usually an instructions to quickly set/clear a bool which is much quicker than the read/modify/write of a single bit. From personal measurements I have seen the read/mod/write method be 5x slower than the single instruction method.

The idea is that one word of memory can be used as an address in the address space (i.e., a word is wide enough to hold a pointer). If an address were larger than a word, addressing would require multiple sequential words. That's not impossible, but is unreasonably complicated (and likely slow).

So, given an w-bit word, how many values can that word represent? How may addresses can it denote? Since any bit can take on two values, we end up with 2·2·2·…·2 = 2^w addresses. Addresses are counted starting by zero, so the highest address is 2^w - 1.

Example using a three-bit word: There are 2³=8 possible addresses:

bin: 000 001 010 011 100 101 110 111
dec:   0   1   2   3   4   5   6   7

The highest address is 2³-1 = 8 - 1 = 7. If the memory is an array of bytes and the address is an index to this array, then using a three-bit word we can only address eight bytes. Even if the array physically holds more bytes, we cannot reach them with a restricted index. Therefore, the amount of virtual memory is restricted by the word size.