MC68000 "size" (was Re: modern serial terminal)

Eric J Korpela korpela at ssl.berkeley.edu
Thu Nov 8 20:41:48 CST 2007


> > Look at the size of the ALU and the accumulator(s)/data registers, then
> > come back to me.
>
> The MC68000 only has a 16 bit ALU and longword operations are done in
> low/high word sequence.

That's exactly what I was going to say.   Because the ALU is 16 bits
and the data bus is 16 bit the optimal word size for handling small
integers and pointer offsets is 16 bits.  When you get to the 68020
the optimal word size for small integers and pointer offsets is 32
bits.  When you need to do things fast, the difference is very
significant.

The 68000 (and the 68010) are 16 bit processors that happen to have 32
bit register groups.  The same way the Z80 is an 8 bit processor that
happens to have 16 bit register groups and the Z8000 is a 16 bit
processor that happens to have 32 bit register groups.  The primary
difference is that Motorola decided not to make the high word
independently addressable which make it look less kludgy.

Somehow people are turning this statement into some sort of praise for
intel's architecture.  It's not.  Intel's segmentation scheme on the
8086 and 80286 sucks.  But just because the 80286 sucks and it's a 16
bit processor, that doesn't turn the 68000 into a 32 bit processor.
It's just a better 16 bit processor.

Eric



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