4 ms·
It was a long time ago (37 years?) but the main difference is this: The Motorola 68000 has 8 32-bit data registers and 8 32-bit address registers. They are nam
by chihuahua 2y ago
It was a long time ago (37 years?) but the main difference is this:
The Motorola 68000 has 8 32-bit data registers and 8 32-bit address registers. They are named D0..D7 and A0..A7. You can use these in most instructions in the obvious/expected way. Maybe it's not completely orthogonal, but close enough. With those registers, you can keep 8 intermediate values and 8 pointers, and perform reasonable operations on them (add this and that, and store at that address)
The 80286 instruction has just 4 16-bit registers, 4 more registers with strange names (BP, SI, DI, SP) and 4 segment registers. None of these are general-purpose. I just looked it up and apparently, you can only use AX and DX for multiply/divide instructions. The segmented addressing is what is most repulsive about the whole thing: you can't just put an address into a register. You have to use a segment register and use an offset so the real address is (segment register value)*16 + offset. The reason for all this bad design is presumably compatibility with even more primitive earlier processors (8086, 8088, 8085, 8080).
All this backward compatibility is so painful and limiting. The 68k was launched in 1979 and it gets all of this right (32-bit addresses, general purpose registers), but the 8086 was launched in 1978 and it's so primitive and ugly.
It reminds me of Windows which even today has certain dumb features such as "\" being the path separator for compatibility with CP/M, some primitive relic of the 1970s. I understand the importance of backward compatibility for commercial success, but it's so ugly and I just don't want to have to deal with it decades later. Yes I know that the NT kernel can theoretically use a different path separator, but in practice, we're all still using \ for paths and / for command line options. Because CP/M was so incredibly great in 1975.