4 ms·

Unfortunately, 6502 can't XOR the accumulator with itself. I don't recall if the Z80 can, and loading an immediate 0 would be most efficient on those anyway.

bonzini 6mo ago
The Z80 can do either LD A,0 or SUB A or XOR A, but the LD is slower due to the extra memory cycle to load the second byte of the instruction.
blywi 6mo ago
XOR A absolutely works on Z80 and it's of course faster and shorter than loading a zero value with LD A,0. LD A,0 is encoded to 2 bytes while XOR A is encoded as a single opcode. XOR A has the additional benefit to also clear all the flags to 0. Sub A will clear the accumulator, but it will always set the N flag on Z80.
- classichasclass 6mo ago
  Ah, thanks, I couldn't recall off the top of my head.
- eichin 6mo ago
  Yeah, the article seems to have missed the likely biggest reason that this is the popular x86 idiom - that it was already the popular 8080/Z80 idiom from the CP/M era, and there's a direct line (and a bunch of early 8086 DOS applications were mechanically translated assembly code, so while they are "different" architectures they're still solidly related.)
- dmitrygr 6mo ago
  should set Z too
repelsteeltje 6mo ago
You're absolutely right, I stand corrected. The 6502 gets by doing immediate load: 2 clock cycles, 2 bytes (frequently followed by single byte register transfer instruction). Out of curiosity I did a quick scan of the MOS 1.20 rom of the BBC micro: LDY #0 (a0 00): 38 hits LDX #0 (a2 00): 28 hits LDA #0 (a9 00): 48 hits
- tom_ 6mo ago
  Are you sure you're not an LLM? There is no way anybody writing 6502 would do anything else, because there's no other way to do it. (You can squeeze in a cheeky Txx instruction afterwards to get a 2-or-more-for-1, if that would be what you need - but this only saves bytes. Every instruction on the 6502 takes 2+ cycles! You could have done repeated immediate loads. The cycle count would be the same and the code would be more general.)
  - repelsteeltje 6mo ago
    > Are you sure you're not an LLM? Hard to tell, but I don't think so ;-) I suppose using Txx instructions rather than LDx is more of an idiom than intended to conserve space. Also, could an LDx #0 potentially be 3 cycles in the edge case where the PC crosses a page boundary? (I'm probably confused? Red herring?)
    - tom_ 6mo ago
      I don't know how the 6502's PC increment actually worked, but it was an exception to the general rule of page crossings (or the possibility thereof) incurring a penalty, or, as was also sometimes the case, just ignored entirely. (One big advantage of the latter approach: doing nothing does take 0 cycles.) The full 16 bits would be incremented after each instruction byte fetched, and it didn't cost any extra if there was a carry out of the MSB.

bonzini 6mo ago

The Z80 can do either LD A,0 or SUB A or XOR A, but the LD is slower due to the extra memory cycle to load the second byte of the instruction.

blywi 6mo ago

XOR A absolutely works on Z80 and it's of course faster and shorter than loading a zero value with LD A,0. LD A,0 is encoded to 2 bytes while XOR A is encoded as a single opcode. XOR A has the additional benefit to also clear all the flags to 0. Sub A will clear the accumulator, but it will always set the N flag on Z80.

classichasclass 6mo ago
Ah, thanks, I couldn't recall off the top of my head.
eichin 6mo ago
Yeah, the article seems to have missed the likely biggest reason that this is the popular x86 idiom - that it was already the popular 8080/Z80 idiom from the CP/M era, and there's a direct line (and a bunch of early 8086 DOS applications were mechanically translated assembly code, so while they are "different" architectures they're still solidly related.)
dmitrygr 6mo ago
should set Z too

repelsteeltje 6mo ago

You're absolutely right, I stand corrected. The 6502 gets by doing immediate load: 2 clock cycles, 2 bytes (frequently followed by single byte register transfer instruction). Out of curiosity I did a quick scan of the MOS 1.20 rom of the BBC micro: LDY #0 (a0 00): 38 hits LDX #0 (a2 00): 28 hits LDA #0 (a9 00): 48 hits

tom_ 6mo ago
Are you sure you're not an LLM? There is no way anybody writing 6502 would do anything else, because there's no other way to do it. (You can squeeze in a cheeky Txx instruction afterwards to get a 2-or-more-for-1, if that would be what you need - but this only saves bytes. Every instruction on the 6502 takes 2+ cycles! You could have done repeated immediate loads. The cycle count would be the same and the code would be more general.)
- repelsteeltje 6mo ago
  > Are you sure you're not an LLM? Hard to tell, but I don't think so ;-) I suppose using Txx instructions rather than LDx is more of an idiom than intended to conserve space. Also, could an LDx #0 potentially be 3 cycles in the edge case where the PC crosses a page boundary? (I'm probably confused? Red herring?)
  - tom_ 6mo ago
    I don't know how the 6502's PC increment actually worked, but it was an exception to the general rule of page crossings (or the possibility thereof) incurring a penalty, or, as was also sometimes the case, just ignored entirely. (One big advantage of the latter approach: doing nothing does take 0 cycles.) The full 16 bits would be incremented after each instruction byte fetched, and it didn't cost any extra if there was a carry out of the MSB.