8051 24bit BCD division

Question

I am currently working on a 8051 assembly program which receives a set of BCD inputs and calculates the mean of this set. However, I couldn't figure out a decent way to divide 24bit BCD number. For instance, say that the sum of the series is 154128 and I am storing this data in 3 bytes: 15H, 41H, 28H and need to divide this by 13. Any ideas how should I proceed? Thanks already.

Answer 1

There are two very simple methods. The simplest is to just successively subtract 13 from 15 41 28 and count the number of times before the result goes negative. It's horribly inefficient but it will eventually yield the correct answer for the integer part of the quotient give or take one. An O(n) algorithm so it can be very slow for large dividend and small divisor.

The second is to use the subtract and shift algorithm you were taught in elementary school. You can do this with BCD numbers. Subtract up to 9x and if the result goes negative then add it back again, so you have as many as 10 operations per digit of result-- so O(log(n)). Pocket calculators will use something like this algorithm. You should be able to design, implement and test this algorithm in a couple hours at most.

A third approach would be to convert the number to binary, do the calculations, then convert back again, which might make sense if there are many calculations to be done, but would be slower than BCD division.