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by sischoel 1462 days ago
Why is that though? Wouldn't something like `mov eax 0` also work?
2 comments

Absolutely (well, if you add the missing comma :) ), just less efficient.

That 0 has to come from somewhere, while in the other case XORing a register with itself does not involve loading any data. It's also shorter.

It begs a question though: How many instructions in <insert ISA here> are equivalent? I assume that a compiler writer has a list of equivalents and will typically choose the shorter one?
For RISC ISAs like MIPS, then you can do a lot of

ADD Reg0 to Reg0 and store result in Rx

OR Reg0 with reg0 and store result in Rx

XOR Reg0 with reg0 and ..

The only issue is that for RISC, all these instructions are of equal length, so flipping them around would gain you very little, or more likely zero effect unless you are chasing some corner case thing like "XOR instruction value compresses slightly better than ADD because.."

There are a number of considerations there. Size is only one of them. Speed and internal processor state effects are two others. For instance, a larger, slower instruction might prevent a pipeline stall in a particular function or might enable loop unrolling or might allow a shorter loop unrolling, while in a similar function that doesn’t pipeline the same way, the compiler will choose a faster instruction.
You're right I completely overlooked pipelining. I guess I'm out of my depth here
Well that 0 that you are loading comes from the instruction, so it is already "there". It boils down to the fact that the instruction is sorter.

In fact in theory the load is slower, because XOR has data dependencies on the arguments. So an out-of-order processor could be delayed. However x86 has special logic that XOR with itself doesn't carry any dependencies on the arguments.

In addition to other concerns, processors usually treat xor specially since it was the best way to zero things for so long that it became ubiquitous. Often its performance impacts are equivalent to a noop.