6 May
2010
6 May
'10
10:34 p.m.
On 5/7/10 1:21 AM, Jim Leonard wrote:
On 5/6/2010 12:41 PM, Tony Duell wrote:
Well you've got to admit, it does depend on how you look at it. The
8088 has an 8-bit data bus and it does two bus cycles to move a 16-bit
value. You know...just like an 8080. ;)
It's just like parallelism. There are many levels of parallelism in
computing. Most of our processors today are "parallel processors" in
that they are bit-parallel, in contrast to the the PDP-8/S for example.
Then there's parallelism at the instruction level (superscalar
execution), parallelism at the processor level (multiprocessor,
multi-core), parallelism at the system level (clusters), etc etc etc.
Above you said "number of bits that can be cahnged in a single
operation". What constitutes a "single operation"? An atomic
instruction? A microinstruction? A micro-op in a PentiumPro? Howabout
SIMD? A SIMD instruction changes LOTS of bits in a single operation,
and in something like a PowerPC, you can determine just how many are
affected. Does this mean it's a variable-"width" processor?
This is just like the tired old debates on what was the first
personal computer and whether an FPGA-based (say) PDP-11 is an emulation
or a hardware processor. It really depends on how/where you look at it
and what context you're speaking in, and there's never a clean ending of
the argument.
-Dave
--
Dave McGuire
Port Charlotte, FL
Is the IBM PC
8 bit (width of the data bus) or 16 bit?
16-bit. Until the day I die.
I thought it was proper practice to note a CPU based on the size of it's
internal registers and/or the number of bits that can be changed in a
single operation.
Is someone seriously challenging that the IBM PC 5150 wasn't a 16-bit
computer?