How much RAM is 64K 36bit words of Core Memory 793

First... how big is your byte? I've worked on architectures where the "byte" was defined as six bits. I've heard of, but never used myself, others where it was nine bits.

Second... what does K mean? 1024 (the usual computer "K", "Kilo") or 1,000? I've seen both.

But, let's buttume you mean eight bits to a byte and 64K is the amount number represented by 16 binary interpreted bits in an unsigned manner. Then direct match is going to say that ((64 * 1024) * 36) 8 = 294912, or 288K bytes. But on such an architecture this may still not be very meaningful since there may be no means direct of addressing that data as eight bit bytes. It works out to 4.5 bytes per word. We've heard of it done (so there's probably someone on this NG who's done it), but it's more likely that such an architecture called a "byte" six bits and stuffed six of them into a word.

As a real world example I worked with a lot, the DG Nova line had 32K (32 * 1024) words of sixteen bits each. In early versions of the architecture there was no direct means to address bytes. But there was a convention for doing it, which was later implemented directly in hardware. Doing text manipulation in FORTRAN on this machine was a wonder to behold. As another example, I worked on something else (the name escapes me now) where the 36 bit words were, as noted above, broken into six bytes, six bits each. This time, however, it was directly supported by hardware. Though the word and byte pointers were not compatible but there was a convention for converting them.

I expect others will chime in with other comments on their experiences.

- Bill