16 bit multiply in PASM - another question

Chris_D

Hello,

I have been working in PASM for a few months now and I really like it.· I can handle the basic stuff pretty good but cannot seem to grasp how to use the example multiplication and division routines referenced in the manual (Appendix B, Math samples and Function tables).

It appears they have to be treated as a subroutine and called which I have not figured out how to do.· Does someone know of an example program that is VERY simple in nature that takes a couple of variables and performs the multiplication on them?·

Ultimately, what I need to do is square two variables.· Both variables will be able to fit in 16 bit vars so it looks like the multiplication routine would be best suited.· However, expecting to do division in PASM at some point in time, I hope to learn how to use both of those routines at the same time.

Thanks

Chris

Post Edited (Chris_D) : 3/12/2010 11:26:14 PM GMT

heater

All you have to do is set up the x and y values that multiply is going to work on and then call it. The result ends up in y.
Of course you have to define the LONGs to hold your numbers and x, y and the temporary variable t that multiply uses.

enter
        mov   x, someNumber_1      'Set x parameter of multiply to some number
        mov   y, someNumber_2      'Set y parameter to some other number
        call  #multiply                         'Do the multiply
        mov   result, y                         'Result is in y

'' Multiply x[noparse][[/noparse]15..0] by y[noparse][[/noparse]15..0] (y[noparse][[/noparse]31..16] must be 0)
' on exit, product in y[noparse][[/noparse]31..0]
'
multiply        shl     x,#16            'get multiplicand into x[noparse][[/noparse]31..16]
                mov     t,#16            'ready for 16 multiplier bits
                shr     y,#1      wc     'get initial multiplier bit into c
:loop if_c      add     y,x       wc     'if c set, add multiplicand to product
                rcr     y,#1      wc     'put next multiplier in c, shift prod.
                djnz    t,#:loop         'loop until done
multiply_ret    ret                      'return with product in y[noparse][[/noparse]31..0]
 
someNumber_1     long      1234
someNumber_2     long      5678
result                        long      0

x             long  0
y             long  0
t             long  0

If you are multiplying by small constant (< 9 bits) value put it directly in the mov instruction e.g.

mov someNumber, #42

Don't forget the example multiply in the manual is for positive numbers only.

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For me, the past is not over yet.

Chris_D

Heater,

Thanks very much for responding.· That is exactly what I needed and more.· I completely forgot about the positive only values.· I suspect that I can work that out in the rest of the code without too much trouble though.

Thanks again!

Chris

Chris_D

I have another question about this multiply routine.· Is there an average number of clock cycles it takes to run?

If an example value is needed and example would be 3500 x 3500

Thanks

Chris

Mike Green

The subroutine takes 216 system clock cycles (12 * 16 + 24) including the call. The other 3 mov's take another 12 clock cycles. The time is not dependent on the values used.

Chris_D

Mike,

Thanks much for answering my question (and the follow up before I even asked it)!

Knowing that the time is a constant helps a lot as I work through a new program I am writing.· It will be a time-critical program and knowing the times of all the "little things" is of great importance.

Thanks again

Chris