3 ms·
80-bit wide registers isn't really arbitrary if you consider that the bulk of the floating point number is a 64-bit significand (and the signifiand ALU makes se
by em3rgent0rdr 14d ago
80-bit wide registers isn't really arbitrary if you consider that the bulk of the floating point number is a 64-bit significand (and the signifiand ALU makes sense as power-of two) and that you don't need as many bits for exponent (it would be wasteful to go to the next power of two up). Memory is stored as 8-bit bytes as the lowest addressable unit, and so the question would be how many extra bytes the number should take, and 80 bits is a nice integer number of 10 bytes.
- mitxela 13d agoThere is no reason that has to be a power of 2.
- dzaima 13d agoExcept the ALU isn't actually 64-bit, it's 67-bit as per article, extra bits for rounding. I'd imagine it was just taken for "prettiness", with 15 bits for exponent being basically reasonable. (maybe some algorithms which double precision per iteration would like it being a power of two? but any such probably vary significantly on initial estimate precision anyway)