July 2002
Intermediate to advanced
320 pages
8h 15m
English
Apparently, many people have made the mistake of assuming that a shift right signed of k positions divides a number by 2k, using the usual truncating form of division [GLS2]. It’s a little more complicated than that. The code shown below computes q = n ÷ 2k, for 1 ≤ k ≤ 31 [Hop].
shrsi t,n,k-1 Form the integer shri t,t,32-k 2**k - 1 if n < 0, else 0. add t,n,t Add it to n, shrsi q,t,k and shift right (signed).
It is branch-free. It also simplifies to three instructions in the common case of division by 2 (k = 1). It does, however, rely on the machine’s being able to shift by a large amount in a short time. The case k = 31 does not make too much sense, because the number 231 is not representable in ...
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