OP is treating XOR as though it were not commutative between registers and memory: > Consequently, there are actually two ways to encode xor eax, eax: ; r

OP is treating XOR as though it were not commutative between registers and memory: > Consequently, there are actually two ways to encode xor eax, eax: ; r/m32, r 31 C0 ; r, r/m32 33 C0 So, what I am saying is that write(memory,register) != write(register,memory) in the byte code, meaning that writing the XOR itself is a noncommutative operation. The machine running the code does however treat both option ultimately to get the same result. The XOR algorithm remains unchanged and remains commutative. The execution routine of the XOR is noncommutative. So, there are two operations here: XOR and executeXOR and OPs article is about using the machine level noncommutativity in the form of different byte strings for executeXOR. The machine still thinks that is is commutative for the end result of the calculation/routine. My comment about cypher XOR plaintext = cyphertext is just a general comment about XOR.