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.
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.
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.
I think the pdp-10 could have 9 bit bytes, depending on decisions you made in the compiler. I notice it's hard to Google information about this though. People say lots of confusing, conflicting things. When I google pdp-10 byte size it says a c++ compiler chose to represent char as 36 bits.
Basically, loading a 0-bit byte from memory gets you a 0. Depositing a 0-bit byte will not alter memory, but may do an ineffective read-modify-write cycle. Incrementing a 0-bit byte pointer will leave it unchanged.
