I have just written the following small paper, which I expect to eventually mount on my page together with C code to implement it. The point is that this is well suited for both conversion on a system with no multiply or divide instructions, or on systems which need to deal with very long binary numbers. I have not dealt with reversing the process, which can easily be done to implement bcd-binary conversions. Meanwhile I would like to hear criticisms as to clarity and general readability, and other things I have not thought of. I expect to edit it accordingly. ________________________________________________________________ An Explanation of the Double-Dabble Bin-BCD conversion Algorithm by C.B. Falconer. ================================================================ The algorithm starts from this description, shifting left, and converting an 8 bit value to BCD. If a prospective BCD digit is five or larger, add three before shifting left. HUNDREDS TENS UNITS BINARY 0000 0000 0000 11111111 Start 0000 0000 0001 11111110 Shift 1 0000 0000 0011 11111100 Shift 2 0000 0000 0111 11111000 Shift 3 0000 0000 1010 11110000 ADD-3 to UNITS 0000 0001 0101 11110000 Shift 4 0000 0001 1000 11110000 ADD-3 to UNITS 0000 0011 0001 11100000 Shift 5 0000 0110 0011 11000000 Shift 6 0000 1001 0011 11000000 ADD-3 to TENS 0001 0010 0111 10000000 Shift 7 0001 0010 1010 10000000 ADD-3 to UNITS 0010 0101 0101 00000000 Shift 8 The rationale for the ADD-3 rule is that whenever the shifted value is 10 or more the weight of that shifted out bit has been reduced to 10 from 16. To compensate, we add 1/2 that 6, or 3, before shifting. We detect the 10 value after shifting by the fact that the value before shifting is 5 or greater. Notice that the rule ensures that the various digits cannot hold non- BCD values. The shifting is the double part of the algorithm. The ADD-3 is the dabble part. It might be better named dabble-double. Alas, history has decreed otherwise. Next, we modify to extract the lsd (least sig. digit) and divide by 10, by adding a shift connection from the units high order bit to the binary low order bit. We also mark the highest order converted bit by x (for 0) and by X (for 1). HUNDREDS TENS UNITS BINARY 0000 0000 x000 11111111 Start 0000 000x 0001 1111111x Shift 1 0000 00x0 0011 111111x0 Shift 2 0000 0x00 0111 11111x00 Shift 3 0000 0x00 1010 11111x00 ADD-3 to UNITS 0000 x001 0101 1111x001 Shift 4 0000 x001 1000 1111x001 ADD-3 to UNITS 000x 0011 0001 111x0011 Shift 5 00x0 0110 0011 11x00110 Shift 6 00x0 1001 0011 11x00110 ADD-3 to TENS 0x01 0010 0111 1x001100 Shift 7 0x01 0010 1010 1x001100 ADD-3 to UNITS x010 0101 -0101 x0011001 Shift 8 | <-- ^ v__________________| Now, try it with a 10 bit number, and move the UNITS register to the right of the BINARY register. Since we no longer have a TENS register we can't to the ADD-3 to it, but we leave it in the annotations for future use. BINARY UNITS 1111111111 x000 Start 111111111x 0001 Shift 1 11111111x0 0011 Shift 2 1111111x00 0111 Shift 3 1111111x00 1010 ADD-3 to UNITS 111111x001 0101 Shift 4 111111x001 1000 ADD-3 to UNITS 11111x0011 0001 Shift 5 1111x00110 0011 Shift 6 1111x00110 0011 ADD-3 to TENS (dummy) 111x001100 0111 Shift 7 111x001100 1010 ADD-3 to UNITS 11x0011001 0101 Shift 8 11x0011001 1000 ADD-3 to UNITS 1x00110011 0001 Shift 9 x001100110 0011 Shift 10 | <-- ^ v_______________| The binary value has become 0x066 or 102 decimal. The units digit is now 3, so the system has divided by 10 and extracted the remainder. The count of shifts is dictated by the length of the binary register, or by when the 'x' marker reaches the left hand bit of the binary register. This is the point at which all the original binary bits have been 'used'. This is sufficient to do BCD conversions, extracting the least significant digit in N operations from an N-bit binary. By repeating it for each digit we end up with roughly N/4 * N operations for the complete conversion, or O(N*N). The next version is identical, but we have broken the binary register up into 4 bit groups. ----BINARY---- THOU HUND TENS UNITS 11 1111 1111 x000 Start 11 1111 111x 0001 Shift 1 11 1111 11x0 0011 Shift 2 11 1111 1x00 0111 Shift 3 11 1111 1x00 1010 ADD-3 to UNITS 11 1111 x001 0101 Shift 4 11 1111 x001 1000 ADD-3 to UNITS 11 111x 0011 0001 Shift 5 11 11x0 0110 0011 Shift 6 11 11x0 0110 0011 ADD-3 to TENS (dummy) 11 1x00 1100 0111 Shift 7 11 1x00 1100 1010 ADD-3 to UNITS 11 x001 1001 0101 Shift 8 11 x001 1001 1000 ADD-3 to UNITS 1x 0011 0011 0001 Shift 9 x0 0110 0110 0011 Shift 10 | <-- ^ v_________________| Now alter the ADD-3 rule to say - Add 3 whenever the digit value is 5 or more, AND the digit is entirely to the right of the x marker, inclusive. Notice that the ADD-3 to TENS suddenly is back in effect. So are two new dabbles, marked by <-- below. ----BINARY---- INDEX THOU HUND TENS UNITS 0 11 1111 1111 x000 Start 1 11 1111 111x 0001 Shift 1 2 11 1111 11x0 0011 Shift 2 3 11 1111 1x00 0111 Shift 3 4 11 1111 1x00 1010 ADD-3 to UNITS 4 11 1111 x001 0101 Shift 4 5 11 1111 x001 1000 ADD-3 to UNITS 5 11 111x 0011 0001 Shift 5 6 11 11x0 0110 0011 Shift 6 7 11 11x0 1001 0011 ADD-3 to TENS (live) 7 11 1x01 0010 0111 Shift 7 8 11 1x01 0010 1010 ADD-3 to UNITS 8 11 x010 0101 0101 Shift 8 9 11 x010 0101 1000 ADD-3 to UNITS 9 11 x010 1000 1000 ADD-3 to TENS <-- 9 1x 0101 0001 0001 Shift 9 10 1x 1000 0001 0001 ADD-3 to HUND <-- 10 x1 0000 0010 0011 Shift 10 | <-- ^ v_________________| and we come out with the BCD conversion to 1023. Notice that all the "ADD-3"s are associated with the following shift. The above has an INDEX column added to correspond. Now we can see when to energize the ADD-3s for each BCD column, in terms of the index value. Units are available immediately, TENS after 4 or more, HUND after 8 or more, and we never can dabble the THOU column. Looks like the rule there should be 12 or more. This bears a startling resemblance to (index DIV 4) indicating a BCD digit. Notice that the conversion has been done in place, with only one 4 bit BCD digit of auxiliary storage. At some size of the binary this will not be enough, but the solution is more zero bits on the left of the binary value. This increases the max index value on conversion. Since the various ADD-3s are limited to a single BCD digit, and they do not interact, they can all be done in parallel, and in fact in parallel with the following shift operation. This means that converting an N bit value to BCD requires no more than N operations, provided that N bit value has sufficient leading zero bits. This is now O(N), which is a large improvement for large binary numbers. A purely software attack will probably not be able to do all the ADD-3s in parallel, but in hardware this is simply a set of gates in the shift path. We can also, if convenient, think of the added units digit as being supplied by an initial 4 bit left rotary shift of the binary register. Problems with software implementation ===================================== We can normally implement left shifts fairly easily in software, with some complications to implement carrying bits across word or octet boundaries. This is probably simplified as much as possible in C by using unsigned chars, and limiting their range to 0..255. Combined with a carry in and a carry out variable, unlimited sized binary shifts can be handled. A practical problem arises with the dabble portion. We could mask off each four bit section of an octet, compare the value to 5, and modify accordingly. This is fairly computationally intensive. Another possibility is to treat the octet as a whole, and pre- prepare a translation table with 256 entries. This should be able to handle the whole octet in one operation. However two table will be needed, depending on whether or not the left hand portion of the octet (see the x bit above) is to be modified. The byte sex of the binary values has to be settled, and it will normally control the order of the BCD values that result. It will usually be convenient for the most significant BCD digit to appear in the lowest address, which in turn implies that the most significant binary bit appears in the lowest address. The reverse is, of course, perfectly feasible. -- Chuck F (cbfalconer@yahoo.com) (cbfalconer@worldnet.att.net) Available for consulting/temporary embedded and systems. <http://cbfalconer.home.att.net> USE worldnet address!

# The double-dabble bin-bcd conversion algorithm

Started by ●March 14, 2004

Posted by ●March 15, 2004

CBFalconer wrote:> I have just written the following small paper, which I expect to > eventually mount on my page together with C code to implement it. > The point is that this is well suited for both conversion on a > system with no multiply or divide instructions, or on systems > which need to deal with very long binary numbers. I have not > dealt with reversing the process, which can easily be done to > implement bcd-binary conversions. > > Meanwhile I would like to hear criticisms as to clarity and > general readability, and other things I have not thought of. I > expect to edit it accordingly.There you are ;-)> The algorithm starts from this description, shifting left, and^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ??? I really don't know what you want to say here. I think my problem is that I don't know what "*The* algorithm" and "*this* description" means.> converting an 8 bit value to BCD. If a prospective BCD digit is > five or larger, add three before shifting left.[...]> The rationale for the ADD-3 rule is that whenever the shifted > value is 10 or more the weight of that shifted out bit has been > reduced to 10 from 16. To compensate, we add 1/2 that 6, or 3, > before shifting. We detect the 10 value after shifting by the > fact that the value before shifting is 5 or greater. Notice > that the rule ensures that the various digits cannot hold non- > BCD values.I know the algorithm and understand why I works. However, I don't think that it is trivial to everyone. Perhaps you should note that the BCD number is doubled by the dabble-bouble. You also should consider proving that the algorithm works by giving a loop-invariant like: BCD-number*256+BINARY = Initial-Binary*2^i (where i is a loop counter)> The shifting is the double part of the algorithm. The ADD-3 is > the dabble part. It might be better named dabble-double. Alas, > history has decreed otherwise.I like this note about the name ;-)> Next, we modify to extract the lsd (least sig. digit) and divide > by 10, by adding a shift connection from the units high order > bit to the binary low order bit. We also mark the highest order > converted bit by x (for 0) and by X (for 1).[...} If one has really unterstood the first variant this should be okay.> Now, try it with a 10 bit number, and move the UNITS register > to the right of the BINARY register. Since we no longer have a > TENS register we can't to the ADD-3 to it, but we leave it in^^ do?> the annotations for future use. > > BINARY UNITS > 1111111111 x000 Start > 111111111x 0001 Shift 1 > 11111111x0 0011 Shift 2 > 1111111x00 0111 Shift 3 > 1111111x00 1010 ADD-3 to UNITS > 111111x001 0101 Shift 4 > 111111x001 1000 ADD-3 to UNITS > 11111x0011 0001 Shift 5 > 1111x00110 0011 Shift 6 > 1111x00110 0011 ADD-3 to TENS (dummy)What's that "dummy"-thing?> 111x001100 0111 Shift 7 > 111x001100 1010 ADD-3 to UNITS > 11x0011001 0101 Shift 8 > 11x0011001 1000 ADD-3 to UNITS > 1x00110011 0001 Shift 9 > x001100110 0011 Shift 10 > | <-- ^ > v_______________|[...]> 11 11x0 0110 0011 ADD-3 to TENS (dummy)see above. [...]> Notice that the conversion has been done in place, with only > one 4 bit BCD digit of auxiliary storage. At some size of the > binary this will not be enough, but the solution is more zero > bits on the left of the binary value. This increases the max > index value on conversion. > > Since the various ADD-3s are limited to a single BCD digit, and > they do not interact, they can all be done in parallel, and in > fact in parallel with the following shift operation. This means > that converting an N bit value to BCD requires no more than N > operations, provided that N bit value has sufficient leading > zero bits. This is now O(N), which is a large improvement for > large binary numbers.Actually, the number of bits needed to store the BCD-representation of a N-bit binary number is O(N).> A purely software attack will probably not be able to do all the > ADD-3s in parallel, but in hardware this is simply a set of > gates in the shift path.I think you should mention the software vs. hardware thing earlier. The (initial) simple double-dabble can already be implemented in O(N) in hardware. The only improvement of the new algorithm is the reduction of necessary registers by a constant factor. Altogether, you seem a bit undetermined: - What is your intended audience? - Do you want to explain *how* the algorithms are implemented or *why* they work? /Jan-Hinnerk PS: Despite all criticism, I was able to learn a few new things from your paper ;-)

Posted by ●March 15, 2004

Jan-Hinnerk Reichert wrote:>... various pertinant criticisms ...> > Altogether, you seem a bit undetermined: > - What is your intended audience? > - Do you want to explain *how* the algorithms are implemented or > *why* they work? > > /Jan-Hinnerk > > PS: Despite all criticism, I was able to learn a few new things > from your paper ;-)Thank-you. Your comments have been saved for a future revision. -- Chuck F (cbfalconer@yahoo.com) (cbfalconer@worldnet.att.net) Available for consulting/temporary embedded and systems. <http://cbfalconer.home.att.net> USE worldnet address!

Posted by ●March 15, 2004

Jan-Hinnerk Reichert schrieb:> CBFalconer wrote: > > Now, try it with a 10 bit number, and move the UNITS register > > to the right of the BINARY register. Since we no longer have a > > TENS register we can't to the ADD-3 to it, but we leave it in > ^^ do? > > > the annotations for future use.> > 1111x00110 0011 ADD-3 to TENS (dummy) > What's that "dummy"-thing?You even annotated the sentence where CBF wrote he'd leave it in; maybe that spelling check blinded you to what the sentence said, because I found that quite clear. I sent my feedback to CBF by email. Michael -- Feel the stare of my burning hamster and stop smoking!

Posted by ●March 15, 2004

A quibble The add three algorithm is just a way of doing BCD math on a binary processor. It's not an inherent part of double-dabble and If you've got a processor that does BCD math you don't need it. I learned double-dabble as a way of doing conversions on paper and there's no reason you can't use double-dabble to convert binary to bases other than ten CBFalconer <cbfalconer@yahoo.com> wrote in message news:<4054AA62.79308D46@yahoo.com>...> I have just written the following small paper, which I expect to > eventually mount on my page together with C code to implement it. > The point is that this is well suited for both conversion on a > system with no multiply or divide instructions, or on systems > which need to deal with very long binary numbers. I have not > dealt with reversing the process, which can easily be done to > implement bcd-binary conversions. > > Meanwhile I would like to hear criticisms as to clarity and > general readability, and other things I have not thought of. I > expect to edit it accordingly. > > > ________________________________________________________________ > An Explanation of the Double-Dabble Bin-BCD conversion Algorithm > by C.B. Falconer. > ================================================================> > The algorithm starts from this description, shifting left, and > converting an 8 bit value to BCD. If a prospective BCD digit is > five or larger, add three before shifting left. > > HUNDREDS TENS UNITS BINARY > 0000 0000 0000 11111111 Start > 0000 0000 0001 11111110 Shift 1 > 0000 0000 0011 11111100 Shift 2 > 0000 0000 0111 11111000 Shift 3 > 0000 0000 1010 11110000 ADD-3 to UNITS > 0000 0001 0101 11110000 Shift 4 > 0000 0001 1000 11110000 ADD-3 to UNITS > 0000 0011 0001 11100000 Shift 5 > 0000 0110 0011 11000000 Shift 6 > 0000 1001 0011 11000000 ADD-3 to TENS > 0001 0010 0111 10000000 Shift 7 > 0001 0010 1010 10000000 ADD-3 to UNITS > 0010 0101 0101 00000000 Shift 8 > > The rationale for the ADD-3 rule is that whenever the shifted > value is 10 or more the weight of that shifted out bit has been > reduced to 10 from 16. To compensate, we add 1/2 that 6, or 3, > before shifting. We detect the 10 value after shifting by the > fact that the value before shifting is 5 or greater. Notice > that the rule ensures that the various digits cannot hold non- > BCD values. > > The shifting is the double part of the algorithm. The ADD-3 is > the dabble part. It might be better named dabble-double. Alas, > history has decreed otherwise. > > Next, we modify to extract the lsd (least sig. digit) and divide > by 10, by adding a shift connection from the units high order > bit to the binary low order bit. We also mark the highest order > converted bit by x (for 0) and by X (for 1). > > HUNDREDS TENS UNITS BINARY > 0000 0000 x000 11111111 Start > 0000 000x 0001 1111111x Shift 1 > 0000 00x0 0011 111111x0 Shift 2 > 0000 0x00 0111 11111x00 Shift 3 > 0000 0x00 1010 11111x00 ADD-3 to UNITS > 0000 x001 0101 1111x001 Shift 4 > 0000 x001 1000 1111x001 ADD-3 to UNITS > 000x 0011 0001 111x0011 Shift 5 > 00x0 0110 0011 11x00110 Shift 6 > 00x0 1001 0011 11x00110 ADD-3 to TENS > 0x01 0010 0111 1x001100 Shift 7 > 0x01 0010 1010 1x001100 ADD-3 to UNITS > x010 0101 -0101 x0011001 Shift 8 > | <-- ^ > v__________________| > > Now, try it with a 10 bit number, and move the UNITS register > to the right of the BINARY register. Since we no longer have a > TENS register we can't to the ADD-3 to it, but we leave it in > the annotations for future use. > > BINARY UNITS > 1111111111 x000 Start > 111111111x 0001 Shift 1 > 11111111x0 0011 Shift 2 > 1111111x00 0111 Shift 3 > 1111111x00 1010 ADD-3 to UNITS > 111111x001 0101 Shift 4 > 111111x001 1000 ADD-3 to UNITS > 11111x0011 0001 Shift 5 > 1111x00110 0011 Shift 6 > 1111x00110 0011 ADD-3 to TENS (dummy) > 111x001100 0111 Shift 7 > 111x001100 1010 ADD-3 to UNITS > 11x0011001 0101 Shift 8 > 11x0011001 1000 ADD-3 to UNITS > 1x00110011 0001 Shift 9 > x001100110 0011 Shift 10 > | <-- ^ > v_______________| > > The binary value has become 0x066 or 102 decimal. The units > digit is now 3, so the system has divided by 10 and extracted > the remainder. The count of shifts is dictated by the length > of the binary register, or by when the 'x' marker reaches the > left hand bit of the binary register. This is the point at > which all the original binary bits have been 'used'. > > This is sufficient to do BCD conversions, extracting the least > significant digit in N operations from an N-bit binary. By > repeating it for each digit we end up with roughly N/4 * N > operations for the complete conversion, or O(N*N). > > The next version is identical, but we have broken the binary > register up into 4 bit groups. > > ----BINARY---- > THOU HUND TENS UNITS > 11 1111 1111 x000 Start > 11 1111 111x 0001 Shift 1 > 11 1111 11x0 0011 Shift 2 > 11 1111 1x00 0111 Shift 3 > 11 1111 1x00 1010 ADD-3 to UNITS > 11 1111 x001 0101 Shift 4 > 11 1111 x001 1000 ADD-3 to UNITS > 11 111x 0011 0001 Shift 5 > 11 11x0 0110 0011 Shift 6 > 11 11x0 0110 0011 ADD-3 to TENS (dummy) > 11 1x00 1100 0111 Shift 7 > 11 1x00 1100 1010 ADD-3 to UNITS > 11 x001 1001 0101 Shift 8 > 11 x001 1001 1000 ADD-3 to UNITS > 1x 0011 0011 0001 Shift 9 > x0 0110 0110 0011 Shift 10 > | <-- ^ > v_________________| > > Now alter the ADD-3 rule to say - Add 3 whenever the digit value > is 5 or more, AND the digit is entirely to the right of the x > marker, inclusive. Notice that the ADD-3 to TENS suddenly is > back in effect. So are two new dabbles, marked by <-- below. > > ----BINARY---- > INDEX THOU HUND TENS UNITS > 0 11 1111 1111 x000 Start > 1 11 1111 111x 0001 Shift 1 > 2 11 1111 11x0 0011 Shift 2 > 3 11 1111 1x00 0111 Shift 3 > 4 11 1111 1x00 1010 ADD-3 to UNITS > 4 11 1111 x001 0101 Shift 4 > 5 11 1111 x001 1000 ADD-3 to UNITS > 5 11 111x 0011 0001 Shift 5 > 6 11 11x0 0110 0011 Shift 6 > 7 11 11x0 1001 0011 ADD-3 to TENS (live) > 7 11 1x01 0010 0111 Shift 7 > 8 11 1x01 0010 1010 ADD-3 to UNITS > 8 11 x010 0101 0101 Shift 8 > 9 11 x010 0101 1000 ADD-3 to UNITS > 9 11 x010 1000 1000 ADD-3 to TENS <-- > 9 1x 0101 0001 0001 Shift 9 > 10 1x 1000 0001 0001 ADD-3 to HUND <-- > 10 x1 0000 0010 0011 Shift 10 > | <-- ^ > v_________________| > > and we come out with the BCD conversion to 1023. Notice that > all the "ADD-3"s are associated with the following shift. The > above has an INDEX column added to correspond. Now we can see > when to energize the ADD-3s for each BCD column, in terms of > the index value. Units are available immediately, TENS after > 4 or more, HUND after 8 or more, and we never can dabble the > THOU column. Looks like the rule there should be 12 or more. > This bears a startling resemblance to (index DIV 4) indicating > a BCD digit. > > Notice that the conversion has been done in place, with only > one 4 bit BCD digit of auxiliary storage. At some size of the > binary this will not be enough, but the solution is more zero > bits on the left of the binary value. This increases the max > index value on conversion. > > Since the various ADD-3s are limited to a single BCD digit, and > they do not interact, they can all be done in parallel, and in > fact in parallel with the following shift operation. This means > that converting an N bit value to BCD requires no more than N > operations, provided that N bit value has sufficient leading > zero bits. This is now O(N), which is a large improvement for > large binary numbers. > > A purely software attack will probably not be able to do all the > ADD-3s in parallel, but in hardware this is simply a set of > gates in the shift path. > > We can also, if convenient, think of the added units digit as > being supplied by an initial 4 bit left rotary shift of the > binary register. > > Problems with software implementation > ===================================== > > We can normally implement left shifts fairly easily in software, > with some complications to implement carrying bits across word > or octet boundaries. This is probably simplified as much as > possible in C by using unsigned chars, and limiting their range > to 0..255. Combined with a carry in and a carry out variable, > unlimited sized binary shifts can be handled. > > A practical problem arises with the dabble portion. We could > mask off each four bit section of an octet, compare the value to > 5, and modify accordingly. This is fairly computationally > intensive. > > Another possibility is to treat the octet as a whole, and pre- > prepare a translation table with 256 entries. This should be > able to handle the whole octet in one operation. However two > table will be needed, depending on whether or not the left > hand portion of the octet (see the x bit above) is to be > modified. > > The byte sex of the binary values has to be settled, and it will > normally control the order of the BCD values that result. It > will usually be convenient for the most significant BCD digit to > appear in the lowest address, which in turn implies that the > most significant binary bit appears in the lowest address. The > reverse is, of course, perfectly feasible.

Posted by ●March 15, 2004

bogax wrote:> > A quibble > > The add three algorithm is just a way of doing BCD math on a > binary processor. It's not an inherent part of double-dabble > and If you've got a processor that does BCD math you don't need > it. I learned double-dabble as a way of doing conversions on > paper and there's no reason you can't use double-dabble to > convert binary to bases other than tenSuch things as the x86 DAA instruction are highly useful as implementation tools, but they are not universally available. BCD math is generally not in favor for various reasons, including memory usage, negation, incompatibility with C standards, etc. Of course it can be used to convert to other bases. In part, that is why I pointed out the division by 10 functional byproduct. Even bases are much more convenient in general. The system is most useful where multiplication and division are not available primitives. -- Chuck F (cbfalconer@yahoo.com) (cbfalconer@worldnet.att.net) Available for consulting/temporary embedded and systems. <http://cbfalconer.home.att.net> USE worldnet address!

Posted by ●March 16, 2004

CBFalconer <cbfalconer@yahoo.com> writes:> bogax wrote: > > > > A quibble > > > > The add three algorithm is just a way of doing BCD math on a > > binary processor. It's not an inherent part of double-dabble > > and If you've got a processor that does BCD math you don't need > > it. I learned double-dabble as a way of doing conversions on > > paper and there's no reason you can't use double-dabble to > > convert binary to bases other than ten > > Such things as the x86 DAA instruction are highly useful as > implementation tools, but they are not universally available. BCD > math is generally not in favor for various reasons, including > memory usage, negation, incompatibility with C standards, etc. > > Of course it can be used to convert to other bases. In part, that > is why I pointed out the division by 10 functional byproduct. > Even bases are much more convenient in general. > > The system is most useful where multiplication and division are > not available primitives.Do you mean binary mul/div? Does anyone have the answer to my question as to how BCD mul/div are implemented on machines with a native BCD math capability?