Putting the chapter together
This capstone combines every idiom from the chapter: a constant-index load, a variable-index lwzx, a neighbor reached from the computed base, a register-register multiply of two loaded values, and a final subtract. Nothing here is new — it is the build-up lessons stacked into one function.
The key efficiency to recognize: the variable index is scaled once with slwi, and that single offset serves both the indexed load lwzx and the neighbor (via add to form the base, then a displacement load). The fixed element at index 0 is a plain lwz 0(r3).
Consider mix(q, j), which combines the first element with the product of two neighbors:
slwi r0, r4, 2 # j * 4
lwz r5, 0(r3) # q[0]
add r4, r3, r0 # r4 = &q[j]
lwzx r3, r3, r0 # q[j]
lwz r0, 4(r4) # q[j+1] (4 / 4 = 1 element past q[j])
mullw r0, r3, r0 # q[j] * q[j+1]
add r3, r5, r0 # q[0] + q[j]*q[j+1]
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Watch how the lone slwi feeds both the lwzx and the add-built base, how mullw multiplies two loaded values (so it is mullw, not mulli), and how the fixed element folds in last. The target assembly has the same ingredients — a fixed element, an indexed element, and a neighbor — but a different arrangement of multiply and subtract. Trace each loaded register from its load to the instruction that consumes it, recover the indices from the displacements, and reconstruct the full expression.
Your task
Write capstone, taking an int* and an int, to reproduce the assembly above.