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

(www.open-std.org)
288 points Twirrim | 4 comments | 17 Oct 24 22:21 UTC | HN request time: 0.43s | source
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pjdesno ◴[17 Oct 24 23:31 UTC] No.41874875[source]▶
During an internship in 1986 I wrote C code for a machine with 10-bit bytes, the BBN C/70. It was a horrible experience, and the existence of the machine in the first place was due to a cosmic accident of the negative kind.
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Isamu ◴[18 Oct 24 03:19 UTC] No.41876076[source]▶
I wrote code on a DECSYSTEM-20, the C compiler was not officially supported. It had a 36-bit word and a 7-bit byte. Yep, when you packed bytes into a word there were bits left over.

And I was tasked with reading a tape with binary data in 8-bit format. Hilarity ensued.

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1. Ballas ◴[18 Oct 24 06:42 UTC] No.41876915[source]▶
That is so strange. If it were 9-bit bytes, that would make sense: 8bits+parity. Then a word is just 32bits+4 parity.
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2. p_l ◴[18 Oct 24 10:01 UTC] No.41877834[source]▶
7 bits matches ASCII, so you can implement entire ASCII character set, and simultaneously it means you get to fit one more character per byte.

Using RADIX-50, or SIXBIT, you could fit more but you'd lose ASCII-compatibility

3. ◴[18 Oct 24 11:01 UTC] No.41878226[source]▶
4. otabdeveloper4 ◴[18 Oct 24 17:04 UTC] No.41881333[source]▶
8 bits in a byte exist in the first place because "obviously" a byte is a 7 bit char + parity.

(*) For some value of "obviously".