From: "DASDBILL2" <dasdbi...@comcast.net> Sent: Thursday, 18 April 2013 1:31 AM
I tried your algorithm with 13 multiplied by 81 and produced the correct answer.
And so it should. It's been around for centuries.
This algorithm is undoubtedly how the microcode for the M (multiply fullword) instruction does its math.
Unlikely. Multiplications by 2 are achieved by a left shift by one place. Typically, the partial product and multiplier are shifted together, thus avoiding division by 2 (i.e. a shift). The m.s. bit of the multiplicand is inspected after each shift; when it is 1, the multiplier is added to the partial product.
