Z80 Divide by 10

It was thus said that the Great dwight elvey once stated: I came across this bit of code from http://www.hackersdelight.org/ to divide by 10: unsigned int div10(unsigned int n) { unsigned int q; unsigned int r; q = (n >> 1) + (n >> 2); q = q + (q >> 4); q = q + (q >> 8); q = q + (q >> 16); q = q >> 3; r = n - ((q << 3) + (q << 1)); return (q + ((r + 6) >> 4); } On the Z80, the (q >> 8) is trivial to handle, and you can probably skip the (q >> 16) statement all-to-gether (since you're only handling 16 bits anyway, this reduces to 0 so the statement there can be skipped). My Z80 skills are pretty weak (and I don't have any references at hand right now) but this looks pretty straight forward to translate. -spc (In fact, the HackersDelight site is quite delightful)