Ok, let's see...

We'll start with 100
value1: 100
value2:

value1: 50
value2: 50

value1: 12
value2: 50

Result = value1 - value2 = -38

The reason this doesn't work is that you cannot add and subtract
divisors like you can add and subtract multipliers.  The equivilant
statement is:

x/8 - x/2 which then becomes
x/8 - 4x/8 which is equal to
-3x/8

The result of which is -3/8 of the original number.  If you add them
instead of subtracting, you get 5/8 of the original number.

The true divide by ten MUST include a factor of 5 in the divisor:
x/(2*5)

Since that cannot be duplicated in rotations (divisions or
multiplications by two) then the PIC can only approximate such a result
with rotations.   As others have said, the approximations involve adding
together smaller divisions of 2:
x/10 is larger than x/16, so add x/16 to the result.  The remainder is
x/26.66.
x/26 is larger than x/32, so add x/32 to the result.  The remainder is x/160
x/160 is larger than x/256, so add x/256 to the result.  The remainder
is x/426.66
etc

As you can see, with just the above three divisions you get 25x/256,
which is fairly close to 1/10.  You can compute the exact error by
subtracting the approximate from the ideal, then dividing by the ideal.
In this case, those three terms give you a 2.3% error.  You may be able
to come closer to the goal in fewer terms by doing 9/10 of x, then
subtracting that from x.

-Adam

Andrew Kieran wrote:

>There are many solutions, but two in particular that I would
>consider:
>
>1)
>o starting with number to be divided in 'value1'
>o Shift right, store result in 'value1' and 'value2'
>o shift right, 'value1' two more times, storing result in
>'value1'
>o subtract 'value2' from 'value1'
>o The result is the original 'value1' divided by 10
>
>This works likes this:
>rrf - divides by two, store in 'value2'
>rrf - now divided by four
>rrf - now divided by eight
>subtract 'value2' - now divided by ten
>
>Note also that the state of the carry flag after the last shift
>can be used to determne whether to round-up the result
>
>2)
>Alterately, I would consider using one of the general purpose
>division routines from the PICLIST.  You should have one for
>your code library anyway.  So unless you're code space is really
>tight and you don't need to do any other division, I'd use the
>general purpose routine rather than this specialized divide by
>ten approach.
>
>That's my $0.02
>
>Andrew
>
>
>
>
>________________________________________________
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>
>---- On    , Phil (phil@SPIDERWEB.COM.AU) wrote:
>
>
>
>>Hello everyone,
>>
>>I'm looking for a divide by 10 routine. The number that I want
>>
>>
>to divide is an
>
>
>>8 bit number but I will very likely need to divide 10 bit
>>
>>
>numbers by ten as
>
>
>>well.
>>
>>While playing with micro processors during the mid 70s I
>>
>>
>remember that
>
>
>>dividing by ten involved a shift right and an add instruction
>>
>>
>but I cannot
>
>
>>remember the sequence.
>>
>>A web search has revealed many 16 bit division routines but
>>
>>
>they look more
>
>
>>complex than I remember for a simple divide by 10.
>>
>>Again, can someone help?
>>
>>--
>>Regards,
>>Phil.
>>
>>--
>>http://www.piclist.com hint: The list server can filter out
>>
>>
>subtopics
>
>
>>(like ads or off topics) for you. See
>>
>>
>http://www.piclist.com/#topics
>
>
>>
>>
>
>--
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>
>
>
>

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