In the loop above, which computation is a candidate for loop-invariant code motion (hoisting out of the loop)? (Assume x and y are not modified in the loop.)

Answer: B. x * y

Loop-invariant code motion moves computations that yield the same value on every iteration in front of the loop so they are computed once. x and y do not change inside the loop, so x * y is the same every time; t = x * y can be computed before the loop and the body becomes a[i] = b[i] * t. b[i] differs for each i, so the whole of b[i] * (x * y) cannot be moved. The assignment to a[i] writes a different element each iteration and stays in the loop.

• A. b[i] changes each iteration, so the whole expression cannot move.
• B. Correct. x * y has the same value on every iteration.
• C. The assignment is needed for each element and stays in the loop.