this post was submitted on 23 Jun 2024
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[–] ulterno@lemmy.kde.social -1 points 5 months ago (1 children)

I see it as the number of possible instructions.

As in, 8 bit 8085 had 2^8^ possible instructions, 32 bit ones had 2^32^ and already had enough possible combinations that we couldn't come up with enough functions to fill the provided space.

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[–] wewbull@feddit.uk 5 points 5 months ago (1 children)

So "instruction encoding length".

I don't think that works though. For something like RISC-V, RV64 has a maximum 32-bit instruction encoding. For x86-64 those original 8-bit intructions still exist, and take up a huge part of the encoding space, cutting the number of n-bit instructions to more like 2^(n-7)

[–] ulterno@lemmy.kde.social 0 points 5 months ago (1 children)

RV64 has a maximum 32-bit instruction encoding

I kinda expected that to happen, since there's already enough to fit all required functions. So yeah, even this is not a good enough criteria for bit rating.

those original 8-bit intructions still exist, and take up a huge part of the encoding space, cutting the number of n-bit instructions to more like 2^(n-7)

err... they are still instructions, right? And they are implemented. I don't see why you would negate that from the number of instructions.

[–] wewbull@feddit.uk 2 points 5 months ago (1 children)

If the 8088 had used all but one 256 8-bit values as legal instructions, all your new instructions after that point would need to start with that unused value and then you can add a maximum of 256 instructions by using the next byte. End result is 511 instructions can be encoded in 16-bits.

[–] ulterno@lemmy.kde.social 0 points 5 months ago

Ah right! I forgot about that.

So you either have to pad all instructions in all previous binaries, or reduce the amount of available instructions in the arch update.