I don't think so--let's consider a simple case, using 8 bits and -10. -10 = 0xf6 Our code looks like: add -10 to dividend skip next instruction if carryout subtract -10 from divident The question is "what is the value of the carry bit here?" Let's try some sample values. 0 + 0xF6 = 0xF6, no carry, subtract 0xF6 = 0, no carry/borrow 9 + 0xF6 = 0xFF, no carry, subtract 0xF6 = 9, no carry/borrow 10 + 0xF6 = 0x00, carry; skip the subtract 11 + 0xF6 = 0x01, carry; skip the subtract So the "ora a" to clear the carry shouldn't be required. Saves a few more cycles. For those porting this method to other platforms, be aware that the 6502/PIC subtract works differently than the x80; the carry bit represents "not borrow" instead of "borrow". The same algorithm can be implemented nearly as efficiently on the 8085 by using the "undocumented" instructions, as it includes a double-precision subtract (as well as a 16-bit arithmetic right shift) as a 1-byte instruction. Note that the BC pair must be used in place of DE to use this instruction. It might be possible to implement the algorithm efficiently on the 8080 by letting BC = -DE and using an ADD HL,BC instead of a SBC HL,DE. One is left without a loop counter register however--but then one could "prime" the A register with a 1 and iterate until the 1 migrates to the sign bit of A, then mask it out as part of the exit code. Cheers, Chuck