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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
by Isamu 2y ago
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.
- bee_rider 2y agoHah. Why did they do that?
- mjevans 2y agoWhich part of it? 8 bit tape? Probably the format the hardware worked in... not actually sure I haven't used real tapes but it's plausible. 36 bit per word computer? Sometimes 0..~4Billion isn't enough. 4 more bits would get someone to 64 billion, or +/- 32 billion. As it turns out, my guess was ALMOST correct https://en.wikipedia.org/wiki/36-bit_computing https://en.wikipedia.org/wiki/36-bit_computing Paraphrasing, legacy keying systems were based on records of up to 10 printed decimal digits of accuracy for input. 35 bits would be required to match the +/- input but 36 works better as a machine word and operations on 6 x 6 bit (yuck?) characters; or some 'smaller' machines which used a 36 bit larger word and 12 or 18 bit small words. Why the yuck? That's only 64 characters total, so these systems only supported UPPERCASE ALWAYS numeric digits and some other characters.
- Ballas 2y agoThat is so strange. If it were 9-bit bytes, that would make sense: 8bits+parity. Then a word is just 32bits+4 parity.
- p_l 2y ago7 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
- deleted 2y ago[deleted]
- otabdeveloper4 2y ago8 bits in a byte exist in the first place because "obviously" a byte is a 7 bit char + parity. (*) For some value of "obviously".