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Why we don't have 128-bit CPUs (www.xda-developers.com)

submitted 4 months ago by [email protected] to c/[email protected]

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[–] [email protected] 143 points 4 months ago* (last edited 4 months ago) (20 children)

We do, depending on how you count it.

There's two major widths in a processor. The data register width and the address bus width, but even that is not the whole story. If you go back to a processor like the 68000, the classic 16-bit processor, it has:

32-bit data registers
16- bit ALU
16-bit data bus
32-bit address registers
24-bit address bus

Some people called it a 16/32 bit processor, but really it was the 16-bit ALU that classified it as 16-bits.

If you look at a Zen 4 core it has:

64-bit data registers
512-bit AVX data registers
6 x 64-bit integer ALUs
4 x 256-bit AVX ALUs
2 x 128-bit data bus to DDR5 (dual edge 64-bit)
~40-bits of addressable physical RAM

So, what do you want to call this processor?

64-bit (integer width), 128-bit (physical data bus width), 256-bit (widest ALU) or 512-bit (widest register width)? Do you want to multiply those numbers up by the number of ALUs in a core? ...by the number of cores on a piece of silicon?

Me, I'd say Zen4 was a 256-bit core, but you could argue any of the above numbers.

Basically, it's a measurement that lost all meaning so people stopped using it.

[–] [email protected] -1 points 4 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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[–] [email protected] 5 points 4 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)

[–] [email protected] 0 points 4 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.

[–] [email protected] 2 points 4 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.

[–] [email protected] 0 points 4 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.

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