On 30/09/2026 22:15, Mark Kahrs via cctalk wrote:
Minor point: The Burroughs B6500/6700 has 51 bits because it had 3 bits of tag on each word. But the underlying CPU was 48 bits.
I looked through https://bitsavers/pdf/burroughs/B6500_6700/1043676_B6500_RefMan_Sep69.pdf and round about p6-5 onwards it goes through a bunch of instruction formats. the tag bits seem to indicate whether this is a Word Data descriptor or Segment Descriptor or a Mark Stack Control Word or whatnot. They look like they are meaningful bits rather than say parity or ECC which are there to catch errors.
It does however say (p2-1 Data Representation) that the "tag bits are inaccessible to normal state (user programs)".
I know nothing about Burroughs other than what I've just read but the tags seem to be an intrinsic part of the processing. It sounds like they are not just internal stuff built by the HW but that a compiler (or whatever created a program on the B6500) had to create the tags as part of that?
Antonio -- Antonio Carlini [email protected]
