如何解释 x86-64 xor 后跟 jle？

Question

If part of the assembly code is as following:

xor %ebp,%ebx
jle some address

does this jle means that it will jump when (%ebx ^ %ebp == 0) because that would set ZF to 1?

Answer 1

That's one of the ways JLE can be true. The other is SF≠ OF , as per the manual: https://www.felixcloutier.com/x86/jcc

Since XOR always clears OF, SF != OF reduces to just SF.

jle after a boolean op will be taken if SF | ZF SF | ZF , ie if the result is <= 0 .

Interesting optimization to avoid test %ebx,%ebx to compare the result against zero ( AND or TEST same,same sets FLAGS identically to cmp reg,0 ).