XOR is left-associative in hardware too
xor has no accumulator register; it always takes exactly two operands. When you chain three or more XOR operands in C, the compiler breaks it into a sequence where each instruction folds in one more value.
Consider xor_four(a, b, c, d), which XORs four values together:
xor r0,r3,r4
xor r0,r5,r0
xor r3,r6,r0
blr
xor r0, r3, r4 computes a ^ b and stores it in r0.xor r0, r5, r0 computes c ^ r0 — that is, c ^ (a ^ b).xor r3, r6, r0 computes d ^ r0 — that is, d ^ (c ^ (a ^ b)).
r3 ends up holding a ^ b ^ c ^ d. Each step feeds its result into the next as one of the two source operands. The intermediate value lives in r0 until the last XOR writes the final result into r3 for the return.
The target assembly is shorter — one fewer xor. Count the xor instructions to determine how many values are being combined, then trace the registers to work out the argument order.
Your task
Write xor_chain, taking three ints, to reproduce the assembly above.