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MitchAlsup1 <mitchalsup@aol.com> wrote:Given 36-bit containers--how do you add 32 or 64-bit constants ??On Mon, 19 Aug 2024 18:52:39 +0000, Brett wrote:>
>MitchAlsup1 <mitchalsup@aol.com> wrote:>On Sun, 11 Aug 2024 0:46:09 +0000, Brett wrote:>
>
>
The thing is that one you go down the GBOoO route, your lack of
registers
"namable in ASM" ceases to become a performance degrader. With renaming
one can have R7 in use 40 times in a 100 instruction deep execution
window.
If this was true we would have 16 or even 8 visible registers, and all
would be fine. x86 does mostly fine with 16, of course x84 had fab and
cubic dollar advantages that dwarfed the register limit.
Careful, here::
>
x86 has LD-OPs and LD-OP-STs which makes the 16 register file feel more
like it has 20-22 registers. Do not underestimate this phenomenon. The
gain from 16-32 registers is only 3%-ish so one would estimate that 22
registers would have already gained 1/2 of all of what is possible.
>64 separate registers was a bridge to far, but it was an interesting>
exercise before it crashed and burned due to the bits being not quite
available. So close, yet so far. I could not make it work.
We remain hobbled by the definition of Byte containing exactly 8-bits.
It is this which drives the 16-bit and 32-bit instruction sizes; and
it is this which drives the sizes of constants used by the instruction
stream.
>
64 registers makes PERFECT sense in a 36-bit (or 72-bit) architecture.
But we must all face facts::
a) Little Endian Won
b) 8-bit Bytes Won
c) longer operands are composed of multiple bytes mostly powers of 2.
d) otherwise it is merely an academic exercise.
>
If you pack 7 instructions in 8 long words that gives you an extra
nibble, > 4 bits.
You can do lots of four operand dual operations, which may get you back
the code density lost, while improving performance.
3 instructions packed in 4 longs gives 64 registers plus four operand{{ note 3 instructions in 4 longs is 85.3-bits per instruction::
dual instructions.
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