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### Network # help on multiplying on 8051

P

#### [email protected]

Jan 1, 1970
0
hi
for a project i should multiply a port content that can be ( 1 to
25000000)
to 85.899346 whit assambly what should i do?
notice that the input can be for example 13543.68 or so ! but ( at
most 2 digit for floating in input so we don't have 23657.687)
any help would be most appriciated

K

#### Keith R. Williams

Jan 1, 1970
0
hi
for a project i should multiply a port content that can be ( 1 to
25000000)
to 85.899346 whit assambly what should i do?
notice that the input can be for example 13543.68 or so ! but ( at
most 2 digit for floating in input so we don't have 23657.687)
any help would be most appriciated

From what you've written it's impossible to tell what you really want
to do. However, if all you want to do is multiply a decimal number by
a constant, just do it as you would by hand; one symbol at a time
(since the 8051 has no decimal multiply). If the multiplicand is in
binary, then you can do it eight bits at a time (symbol = 8 bits).

P

#### Peter Bennett

Jan 1, 1970
0
hi
for a project i should multiply a port content that can be ( 1 to
25000000)
to 85.899346 whit assambly what should i do?
notice that the input can be for example 13543.68 or so ! but ( at
most 2 digit for floating in input so we don't have 23657.687)
any help would be most appriciated

It is not at all clear what you are trying to do. An 8051 port only
has 8 bits, so can only represent numbers from 0 - 255 (or -128 to
+127). Even using two ports, you can only count to 65535.

For anyone to be able to give a useful answer, you would need to
describe the internal (binary) format that you are using for floating
point numbers.

--
Peter Bennett, VE7CEI
peterbb (at) interchange.ubc.ca
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter

J

#### John Jardine

Jan 1, 1970
0
hi
for a project i should multiply a port content that can be ( 1 to
25000000)
to 85.899346 whit assambly what should i do?
notice that the input can be for example 13543.68 or so ! but ( at
most 2 digit for floating in input so we don't have 23657.687)
any help would be most appriciated

The question's a bit unclear.
But, for multiplying two numbers with fixed decimal precisions it's quite
easy to pretend the number(s) are whole integer values and write an integer
"n byte" multiply routine that that will handle the number of bits that can
result. This result is then passed onto another routine that integer divides
the previous answer by the 10 or 100 or 1000 ect that the original was
changed by. The result (integer bits and 'remainder' bits) is then converted
back to BCD, decimal, ascii etc.

Eg The port number might be "12,345,678.90". You know there's only ever two
decimal digits, so call that number "1234567890". (ie. pretend it is 100
times bigger.). The number can now be stored in memory as 4 bytes (32 bits
of binary).
Now pretend the multiplier is "85899346" (ie a million times bigger). Store
this also as 4 bytes of memory.

Code up or find a "32 bit x 32 bit" multplier routine. Use this to multiply
your two 32 bit numbers. The routine will result in an 8 byte (64 bits)

This 8 byte answer now needs dividing back down by "100" and "1000000" (ie
divide the 64 bit (binary) answer by 100 million).
Write or find a "64 bit by 32 bit" divide routine.
Convert the "100 million" into into a 4 byte (32 bit) binary number. (this
value will always be the same for each calculation). Do the division of
your 64 bit and 32 bit numbers.
This is the final answer (in binary of course). So convert this number
(integer part plus the 'remainder' part ) back into BCD etc.

The 'n' byte multiply and divide routines are quite small, understandable
and are standard use for machine code programming. A search on the web
should turn up lots of examples.

regards
john

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