Meaning of "architecture width" - Re: 68K Macs with MacOS 7.5 still in production use...

In a talk on the HP9810 desktop calculator I said, somewhat tongue-in-cheek that it could claim to be 1, 3, 4, 6, 8 or 16 bits. The justifications for those are as follows : The processor has 16 bit registers. It's bit serial, so the normal internal ALU and data path width is 1 bit. But it does BCD additions between the A (accumulator) and T (memory data) registers a nybble (4 bits) at a time. The memory system is odd, with hardware translating a read or write into 2 cycles on successive locations. One location stores the odd bits, the other the even bits. So the physical width of memory for things like the system stack is 8 bits. Even odder, the machine stores user programs as sequences of 6 bit keycodes. User program memory is logically 6 bits wide. But because of the memory control system, such memory is physically 3 bits wide, user program memory expansion consists of Intel 1103 (1K*1bit) DRAMs fitted in sets of 3. So : 1 bit -- physical width of binary ALU and data paths 3 bits -- physical width of user program memory 4 bits -- physical width of BCD ALU and data paths 6 bits -- logical width of user program memory 8 bits -- physical width of data memory 16 bits -- logical width of data memory, physical width of CPU registers Call it what you will :-) -tony