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.
I gave up trying to figure out what the “bitness” of CPUs were around the time the Atari Jaguar came out and people described it as 64 bit because it had 32 bit graphics chip plus a 32 bit sound chip.
It’s been mostly marketing bollocks since forever.
At less than a tenth the size, this is actually a better explanation than the article. Already correcting the fact that we do at the very beginning.
If you absolutely had to put a bit width on the Zen 4, the 2x128 bit data bus is probably the best single measure totaling 256 bit IMO.
Even then, at what point do you measure it? DDR interface is likely very much narrower than the interfaces between cache levels. Where does the core end and the memory begin?
Yes you are 100% right, and I did consider level 3 cache as a better measure, because that allows communication between cores without the need to go through RAM, and cache generally has a high hit rate. But this number was surprisingly difficult to find, so I settled on the data bus.
Anyways it would be absolutely fair to call it 256bit by more than one measure. But for sure it isn’t just 64 bit, because it has 512 bit instructions, so the instruction set isn’t limited to 64 bit. Even if someone was stubborn enough to claim the general instruction set is 64 bit, it has the ability to decode and execute 2 simultaneous 64 bit instructions per core, making at least 128 bit by any measure.
As in, 8 bit 8085 had 28 possible instructions, 32 bit ones had 232 and already had enough possible combinations that we couldn’t come up with enough functions to fill the provided space.
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)
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.
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.
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:
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:
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.
I gave up trying to figure out what the “bitness” of CPUs were around the time the Atari Jaguar came out and people described it as 64 bit because it had 32 bit graphics chip plus a 32 bit sound chip.
It’s been mostly marketing bollocks since forever.
The Jaguar lied with the truth, and I say this as someone who still owns one.
At less than a tenth the size, this is actually a better explanation than the article. Already correcting the fact that we do at the very beginning.
If you absolutely had to put a bit width on the Zen 4, the 2x128 bit data bus is probably the best single measure totaling 256 bit IMO.
Even then, at what point do you measure it? DDR interface is likely very much narrower than the interfaces between cache levels. Where does the core end and the memory begin?
Yes you are 100% right, and I did consider level 3 cache as a better measure, because that allows communication between cores without the need to go through RAM, and cache generally has a high hit rate. But this number was surprisingly difficult to find, so I settled on the data bus.
Anyways it would be absolutely fair to call it 256bit by more than one measure. But for sure it isn’t just 64 bit, because it has 512 bit instructions, so the instruction set isn’t limited to 64 bit. Even if someone was stubborn enough to claim the general instruction set is 64 bit, it has the ability to decode and execute 2 simultaneous 64 bit instructions per core, making at least 128 bit by any measure.
Not to mention most “8-bit” CPUs had a 16 bit address bus.
Yes, because 256 memory locations is a bit limiting.
So, you’re saying it already goes to ‘11’?
I’m surprised some marketing genius at the intel/amd hasnt started using the bigger numbers
I expect the engineers are telling the marketing people “No! You can’t do that. You’ll scare everyone that it’s incompatible.”
32bits is compatible with 64bits, why wouldn’t 128 bits be too?
64bit cut out 16bit compatibility. So I’m guessing the fear is that 128 would cut 32.
Very well said. I think you make your point quite effectively!
I see it as the number of possible instructions.
As in, 8 bit 8085 had 28 possible instructions, 32 bit ones had 232 and already had enough possible combinations that we couldn’t come up with enough functions to fill the provided space.
CC BY-NC-SA
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)
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.
err… they are still instructions, right? And they are implemented. I don’t see why you would negate that from the number of instructions.
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.
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.