16 bit multiply in PASM - another question

Comments

• heater Posts: 3,370

  2010-03-10 14:22 edited 2010-03-10 14:22

  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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• Chris_D Posts: 305

  2010-03-10 15:27 edited 2010-03-10 15:27

  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 Posts: 305

  2010-03-12 23:23 edited 2010-03-12 23:23

  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 Posts: 23,101

  2010-03-12 23:56 edited 2010-03-12 23:56

  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 Posts: 305

  2010-03-13 00:05 edited 2010-03-13 00:05

  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