alt Hacker NewsI remember a lot of code zeroing registrers, dating at least back from the IBM PC XT days (before the 80286).
If you decode the instruction, it makes sense to use XOR:
- mov ax, 0 - needs 4 bytes (66 b8 00 00) - xor ax,ax - needs 3 bytes (66 31 c0)
This extra byte in a machine with less than 1 Megabyte of memory did id matter.
In 386 processors it was also - mov eax,0 - needs 5 bytes (b8 00 00 00 00) - xor eax,eax - needs 2 bytes (31 c0)
Here Intel made the decision to use only 2 bytes. I bet this helps both the instruction decoder and (of course) saves more memory than the old 8086 instruction.
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> - mov ax, 0 - needs 4 bytes (66 b8 00 00) - xor ax,ax - needs 3 bytes (66 31 c0)
You don't need operand size prefix 0x66 when running 16 bit code in Real Mode. So "mov ax, 0" is 3 bytes and "xor ax, ax" is just 2 bytes.
> If you decode the instruction, it makes sense to use XOR:
> - mov ax, 0 - needs 4 bytes (66 b8 00 00) - xor ax,ax - needs 3 bytes (66 31 c0)
Except, apparently, on the pentium Pro, according to this comment: https://randomascii.wordpress.com/2012/12/29/the-surprising-..., which says:
“But there was at least one out-of-order design that did not recognize xor reg, reg as a special case: the Pentium Pro. The Intel Optimization manuals for the Pentium Pro recommended “mov” to zero a register.”
I don't know enough of the 8086 so I don't know if this works the same, but on the Z80 (which means it was probably true for the 8080 too), XOR A would also clear pretty much all bits on the flag register, meaning the flags would be in a known state before doing something that could affect them.
> the IBM PC XT days (before the 80286)
Fun fact - the IBM PC XT also came in a 286 model (the XT 286).
As the author says, a couple of extra bytes still matter, perhaps more than 20ish years ago. There are vast amounts of RAM, sure, but it's glacially slow, and there's only a few tens of kBs of L1 instruction cache.
Never mind the fact that, as the author also mentions, the xor idiom takes essentially zero cycles to execute because nothing actually happens besides assigning a new pre-zeroed physical register to the logical register name early on in the pipeline, after which the instruction is retired.