是MOV％ESI，ESI％无操作或不x86-64的？

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问题描述

我有点被注释为Linux内核，arch/x86/include/asm/nops.h.它指出，

I guess the author implied the machine instructions ('89 F6' and '8D 76 00', respectively) there rather than assembly instructions. It follows from the description of LEA in Intel Software Developer's Manual Vol 2A that the latter instruction (lea 0x00(%rsi), %esi) does the same as the the former, mov %esi,%esi.

So this reduces to the question, whether mov %esi,%esi is actually a no-op on x86-64.

mov does not change flags. This kind of mov does not change memory either. It seems, if it changes something besides %rip, that should be general purpose registers. But I have no clue how it can change the contents of %rsi or whatever. If you manipulate the lower half of a general purpose register, the upper half should not change, right?

解决方案

mov %esi, %esi

zeros out the high 32 bits of %rsi, and is therefore not a no-op on x86_64.

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