←back to thread

C++ proposal: There are exactly 8 bits in a byte

(www.open-std.org)
182 points Twirrim | 3 comments | 17 Oct 24 22:21 UTC | HN request time: 0s | source
Show context
donatj ◴[17 Oct 24 23:55 UTC] No.41875031[source]▶
So please do excuse my ignorance, but is there a "logic" related reason other than hardware cost limitations ala "8 was cheaper than 10 for the same number of memory addresses" that bytes are 8 bits instead of 10? Genuinely curious, as a high-level dev of twenty years, I don't know why 8 was selected.

To my naive eye, It seems like moving to 10 bits per byte would be both logical and make learning the trade just a little bit easier?

replies(6): >>41875041 #>>41875052 #>>41875110 #>>41875147 #>>41875204 #>>41875211 #
dplavery92 ◴[17 Oct 24 23:56 UTC] No.41875041[source]▶
Eight is a nice power of two.
replies(1): >>41875063 #
donatj ◴[18 Oct 24 00:01 UTC] No.41875063[source]▶
Can you explain how that's helpful? I'm not being obtuse, I just don't follow
replies(4): >>41875100 #>>41875101 #>>41875319 #>>41875462 #
1. spongebobstoes ◴[18 Oct 24 00:06 UTC] No.41875101[source]▶
One thought is that it's always a whole number of bits (3) to bit-address within a byte. It's 3.5 bits to bit address a 10 bit byte. Sorta just works out nicer in general to have powers of 2 when working on base 2.
replies(1): >>41875607 #
2. cogman10 ◴[18 Oct 24 01:37 UTC] No.41875607[source]▶
This is basically the reason.

Another part of it is the fact that it's a lot easier to represent stuff with hex if the bytes line up.

I can represent "255" with "0xFF" which fits nice and neat in 1 byte. However, now if a byte is 10bits that hex no longer really works. You have 1024 values to represent. The max value would be 0x3FF which just looks funky.

Coming up with an alphanumeric system to represent 2^10 cleanly just ends up weird and unintuitive.

replies(1): >>41876097 #
3. Spivak ◴[18 Oct 24 03:24 UTC] No.41876097[source]▶
We probably wouldn't have chosen hex in a theoretical world where bytes were 10 bits, right? It would probably be two groups of 5 like 02:21 == 85 (like an ip address) or five groups of two 0x01111 == 85. It just has to be one of its divisors.