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17-02-88/T\OSB13

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File contents
OSBITS - An Exploration of the BBC Micro at Machine Level

By Programmer

...........................................................


Part 13: Multi-Byte Multiplication

While I have no desire to labour the point it is worth
taking a detailed look at multiplication where the numbers
being multiplied are both several bytes long.  I have called
this multi-byte arithmetic although it is often known as
multi-precision arithmetic.

The last module included a routine which multiplied a large
number by a small one; a four byte number by a one byte
number.  The extension of this to arbitrarily large numbers
is straightforward but I want to deal with it separately so
that any confusion that may still be in your mind from last
time is dispelled.

Let's recap on the method employed in multiplication.  It is
exactly like the long multiplication you will have done at
school (and probably not used since) except, of course, that
we will be working in binary.  Ironically this makes it
easier although when working on paper we have to be careful
about binary maths because we are unused to it.

Returning to the single byte example from last time we have
the two numbers 11 and 27 which we want to multiply
together.  11 is 1011 in binary because it is made up of
2^3 + 2^1 + 2^0 (8+2+1) and 27 is 11011 because it is made
up of 2^4 + 2^3 + 2^1 + 2^0 (16+8+2+1).  It is irrelevant
that the numbers are related in a way that is obvious in
binary.

                     11011
                         x
                      1011
                 ---------
                     11011
                    11011o
                  11011ooo
                 ---------
                 100101001

This is how we carried out the multiplication in the last
module.  When we start multiplying large numbers together
though we run into a problem.  If you multiply a number with
x bits by a number with y bits the product has x+y bits (I'm
assuming the numbers are both positive).  If you want to
make sure that your result does not overflow you have to
allow for a result of that size.  This in fact means that
when you do your multiplication you have to have a partial
product workspace that is x+y bits long.  Obviously the more
bits in your partial product workspace the slower the
routine becomes because you are carrying out additions on
more bytes than you want the result to have.

It's worth looking at a different algorithm since there are
several ways of implementing multiplication. First I'll
rewrite the sum above to separate out the two additions.

                     11011
                         x
                      1011
                 ---------
                     11011
                    11011o
                 ---------
                   1010001
                  11011ooo
                 ---------
                 100101001

What we are doing here is shifting the multiplicand left and
then conditionally adding it onto the partial product.  As
an alternative how about shifting the partial product right
instead?  This would look like this:

       11011
           x
        1011
   ---------
       00000       ppws is empty to start with

       11011       lsb of 1011 is 1 so add 11011 to ppws
       11011       new ppws
        1101 1     shift ppws right one bit

       11011       bit 1 of 1011 is 1 so add 11011 to ppws
      101000 1
       10100 01    shift ppws right one bit

       00000       bit 2 of 1011 is 0 so don't add
       10100 01    new ppws
        1010 001   shift ppws right one bit

       11011       bit 3 of 1011 is 1 so add 11011 to ppws
      100101 001   new ppws
       10010 1001  shift ppws right one bit

If we count leading zeros in our multiplier and so carry out
eight operations (the next byte size up) we carry out
another four shifts and get the result

                  1 00101001

On each occasion we are shifting the multiplier right to
start with and then immediately afterwards we shift the ppws
right.  Now this could be a useful feature because if we can
combine the two shifts, multiplier and ppws, into a single
shift operation we will end up with the result in the space
where the multiplier was.

I haven't worried too much about speed before.  With the
number translation program last time the speed of the user
was the limiting factor anyway, so speed considerations were
irrelevant.  With a multiplication speed is a very important
consideration so let's look at the algorithm we have come up
with now.

    Clear ppws (partial product work space)
    Shift multiplier right, puts lsb into carry flag
    If carry set add multiplicand to ppws
    Shift ppws and multiplier right
    Repeat last two lines until all bits processed

When you have finished the lower bytes of the result have
replaced the multiplicand and the 'overflow' remains in the
ppws.  Now that looks short and to the point, the only
asymmetrical part being the initial shift, but we've got to
start somewhere.

So let's turn it into a program.

When rotating a multi-byte number you start at the far end,
(i.e. if rotating left start at the right and vice versa)
and use ASL or LSR as the first rotation so that you do not
shift the carry value in, because you do not want it!  The
other rotates will be ROL or ROR because then you use the
carry to carry the relevant bit over from one byte to the
next.  In this case we will treat the ppws and multiplier as
a single long block of memory with the ppws to the left of
the multiplier.

There is no limit to the size of numbers you can multiply
together with this method.  Remember that a four byte number
multiplied by another four byte number can result in an
eight byte result, so you have to allow for this
possibility.  If we were dealing solely with positive
numbers then, by treating the ppws and multiplier as a
single piece of work space, we would automatically have
enough room for a large result.  We could test for an
overflow by looking to see if anything was in the top 4
bytes.

A negative number with 4 bytes is defined as one with the
top most bit (bit 31) set.  Clearly if we are working to a
64 bit result then that number should be held in memory in
8 bytes with the top bit (number 63) set.  All the bits
between 31 and 63 would also be set.  To make this routine
operate totally correctly for negative numbers we would have
to expand the multiplier and multiplicand into 64 bit
numbers and have a routine that ran through 64 times instead
of 32.  For simplicity I have not done so.  The result is
that it is possible to generate a result (positive or
negative) that is too big but it is very difficult to trap.

The simplest solution would be to check to see what sign the
result should be and then convert both numbers to positive,
carry out the multiplication, and convert back afterwards if
needed.  Then you could just check to see if anything was
left in ppws after the multiplication.  Anything there would
indicate an overflow and hence an error.  Although we didn't
error trap the addition and subtraction in module 2 it could
have easily been done by branching to the error code
whenever the overflow flag was set after adding or
subtracting the highest byte.

The input routine in the last module can be modified to
always give a positive result and an indication of sign.
I am not going to use that input module here because all the
superfluous code is likely to cloud the main issue, which is
the multiplication.  I leave it to you to bolt the routines
together.  So I use dear old BASIC indirection operators to
put the numbers to be multiplied into the workspace and to
read the result out afterwards.

The multiplicand goes into 'multiplicand' and the multiplier
goes into 'multiplier'.  'ppws' is the partial product
workspace.  This arrangement fits nicely with the
requirement for ppws and multiplier to be together.  [In
fact the memory used for multi-byte maths does not need to
be contiguous, in a block, but it is sensible to have it
so.]

A further note on speed.  The 6502 works faster when it is
using zero page workspace, so we should really use some of
this for our routines.  I am not going to in this example
because it keeps a consistency with earlier modules and I
leave it to you to make the adjustment if you want to.  It
doesn't affect the algorithm at all.

Firstly the workspace is cleared and then a counter is
initialised to the number of bits in the numbers.  Since
this routine will cope with negative numbers as easily as
positive ones (2's complement negative that is) we will need
all 32 bits.  -1 is &FFFFFFFF (I will refrain from printing
out 32 '1's just to illustrate this point.)

The first thing that happens in the code is that we clear
our partial product workspace and set the bit counter to 32
(or however many bits you are working with).  To get our
algorithm ready we have to first rotate or shift the
multiplier right to put its least significant bit (lsb) into
the carry flag.  We can then use branches on carry to decide
whether or not to add the multiplicand onto the contents of
the partial product workspace.

'mbm_loop' (for multi-byte multiplication) is the top of the
main loop.  In it we first check the carry flag to see if
the lsb of the multiplier was set.  If it was then we will
add the multiplicand onto the ppws but if not the program
goes to 'no_add_mbm' where both the ppws and the multiplier
are rotated right.  (Where I refer to 'ws_something' I mean
the group of bytes that make up that word, in this case
'multiplicand' to 'multiplicand+3'.)  This rotation puts the
lsb of the multiplier into the carry flag ready for another
run around the loop.

Remember we rotate the ppws and multiplier each time we go
around the loop whether or not we add the multiplicand to
the partial product.  Following the rotation the X register,
which is counting the loop for us, is reduced by one. 
Counting down is usually better than counting up because a
simple BNE will suffice to trap zero rather than having to
CMP with anything.  Here we BNE back to 'mbm_loop'.

When we finally fall through the BNE we reach an RTS which
finishes the routine.  The four byte result is in
'multiplier' with any overflow in 'ppws'.  Then follow the
three FNEQUM functions that reserve memory for the
workspace.  If you have BASIC 2 or later you can replace
them with EQUD 0 in each case.

The BASIC at the end prints out the result with a check and
tells you what exactly is in 'ppws' and 'multiplier' so you
can see what happens if you enter large numbers.  However,
for a product under 32 bytes the results will be correct for
positive and negative.

Multiplication and division routines are amongst the most
time sensitive in maths.  In an early module when I said
that you might wish to do more than just multiply faster I
was in fact hiding the truth.  The routines in BBC BASIC are
so efficient that you can multiply two integers together
faster from his BASIC than with this code here.  However I
am aiming for clarity (built for comfort not for speed I
believe) which is my excuse and I will stick to it.

If you wish to expand (or contract) this routine to work
with a different number of bytes you simply change the size
of the workspace areas accordingly and add or subtract from
the lists of LDA/ADC/STAs and LSR/RORs.  Remember that by
convention with the 6502 a multi-byte number is stored with
its lowest significant byte in the lowest memory location
and the most significant byte in the highest location.  This
results from the way a 16 bit address must be stored with
the less significant byte first.

Next time we move on to division and an output routine which
needs division to operate.  Then on to a general division
routine similar to the multiplication one in this module.
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00000020  42 43 20 4d 69 63 72 6f  20 61 74 20 4d 61 63 68  |BC Micro at Mach|
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00000040  6f 67 72 61 6d 6d 65 72  0d 0d 2e 2e 2e 2e 2e 2e  |ogrammer........|
00000050  2e 2e 2e 2e 2e 2e 2e 2e  2e 2e 2e 2e 2e 2e 2e 2e  |................|
*
00000080  2e 2e 2e 2e 2e 0d 0d 0d  50 61 72 74 20 31 33 3a  |........Part 13:|
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000005e0  20 20 20 20 20 20 20 20  20 20 20 20 20 20 20 20  |                |
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00000600  20 20 20 20 20 20 20 20  20 2d 2d 2d 2d 2d 2d 2d  |         -------|
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000029f0  20 77 68 65 6e 20 49 20  73 61 69 64 0d 74 68 61  | when I said.tha|
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00002d50  0d 4e 65 78 74 20 74 69  6d 65 20 77 65 20 6d 6f  |.Next time we mo|
00002d60  76 65 20 6f 6e 20 74 6f  20 64 69 76 69 73 69 6f  |ve on to divisio|
00002d70  6e 20 61 6e 64 20 61 6e  20 6f 75 74 70 75 74 20  |n and an output |
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00002da0  6f 70 65 72 61 74 65 2e  20 20 54 68 65 6e 20 6f  |operate.  Then o|
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00002dc0  69 76 69 73 69 6f 6e 0d  72 6f 75 74 69 6e 65 20  |ivision.routine |
00002dd0  73 69 6d 69 6c 61 72 20  74 6f 20 74 68 65 20 6d  |similar to the m|
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00002e00  2e 0d                                             |..|
00002e02
17-02-88/T\OSB13.m0
17-02-88/T\OSB13.m1
17-02-88/T\OSB13.m2
17-02-88/T\OSB13.m4
17-02-88/T\OSB13.m5