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Comment by mytailorisrich | Hacker News Reader

xor reg, reg is indeed the standard way to zero out a register in x86 assembly (it's not just a compiler trick) as it is both shorter and faster than loading the register with zero via a mov.

Apart from that, I'd say that a common use of XOR operations in general are interactions with hardware peripherals where manipulating bit fields are needed.

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Why does mov'ing a constant end up having overhead?

It's not overhead, it's about dependency breaking. 32-bit xors on a single register are universally recognized as a zeroing idiom, which means the CPU doesn't have to wait for the results of previous operations in order to set the value of the applicable register to zero.

In modern CPUs zero'ing idioms aren't even executed, they only get as far as the register allocater. The register allocater will allocate a zero'd physical register for the architectural register that had the idiom applied to it and the job is done.

Could you share a source about register allocation describing optimizations such as the one you described?

You might find what you are looking for by googling "register renaming"

The magic words for this is "zeroing idiom".

It's larger, because it needs to fit a 32bit value of 0 in the instruction, and thus e.g. on x86 needs 5 bytes, whereas xor reg,reg needs 2. As such it was a common code size optimization, which in turn has lead to CPU manufacturers optimizing their CPUs to recognize it and treat it even more efficiently.

A comprehensive answer:

https://stackoverflow.com/questions/33666617/what-is-the-bes...

Basically, xor leads to smaller code and more efficient use of resources.

IIRC immediate operands have to be the same size as the destination, so to zero a 32-bit register you need a 32-bit constant, and it simply makes the whole instruction larger than a simple xor.

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