> From: John Dammeyer <johnd@ISLANDNET.COM>
> >[cut]
> >DAA is a bit tekno for this task.  My philosophy when it comes to
> >binary to decimal decoding is that decimal is almost always used
> >by humans (or printers) and hence speed is not an overriding concern.
> >I would go for the most compact code.  Successive subtraction of
> >powers of ten would be my first choice.  Actual division is
> >certainly unnecessary.  Converting an unsigned 16-bit number to
> >5-digit decimal requires a maximum of 27 2-byte subtractions,
> >20 1-byte subtractions and 5 additions.
>
> Please do post it.  ie: Put your opcodes down as evidence.

So you want me to put my code where my mouth is?  Well I've never
needed to actually do this (let alone on a '57, since I've always
used the '74 and '84).  Time does not permit me to give a real
implementation so you will have to be satisfied with pseudo C-code.
This code would translate in a fairly straightforward way into
16C57 code.  Obviously, some of the code could be 'commoned'.
Contrary to my previous post this will take a worst case, for
conversion of the number 5999x, of 26 2-byte subs, 10 1-byte subs,
3 2-byte adds and 2 1-byte adds -- not counting the digit increments.
The subtractions could be replaced by additions of -ve numbers if
that makes it easier.

The conversion procedure is broken up into very small bites so
that your non-interrupt-capable processor will have only a small
latency between checking for I/O etc.



enum state { start, finish, sub10k, sub1k, sub100, sub10 }
int binary
int divisor
char digit[5]

void bin2dec()
{
  switch (state)
  {
    case start:
      state = sub10k
      digit[0] = digit[1] = ... = digit[4] = '0'
      divisor = 10000
      return
    case sub10k:
      binary -= divisor
      if (binary < 0) [i.e. MSB of binary changed from 0 to 1]
      {
        binary += divisor
        state = sub1k
        divisor = 1000
      }
      else
        digit[0]++
      return
    case sub1k:
      binary -= divisor
      if (binary < 0)
      {
        binary += divisor
        state = sub100
        divisor = 100
      }
      else
        digit[1]++
      return
    case sub100:
      binary -= divisor
      if (binary < 0)
      {
        binary += divisor
        state = sub10
        divisor = 10
      }
      else
        digit[2]++
      return
    case sub10:
      binary -= divisor [S.B. - single byte op]
      if (binary < 0)
      {
        binary += divisor [S.B.]
        state = finish
        digit[4] += binary [S.B.]
      }
      else
        digit[3]++
      return
  }
}


main()
{
  ...
  binary = [binary value to decode]
  state = start
  while (state != finish)
  {
    bin2dec()
    [do something else e.g. service I/O]
  }

  print("decimal = %c%c%c%c%c", digit[0], digit[1], ..., digit[4])

}


Regards,
SJH
Canberra, Australia