HP Forums

For positive integer input the programme returns the original input and the check digit.

eg For input

18081953

the programme returns

18081953
4

For more info on IBM check digit see

http://en.wikipedia.org/wiki/Luhn_algorithm

Edit: Programme faulty, see next posting.

Code:

::

  CK1&Dispatch

  # FF

  ::

    DUP

    {

      BINT0

      BINT2

      BINT4

      BINT6

      BINT8

      BINT1

      BINT3

      BINT5

      BINT7

      BINT9

    }

    SWAPDUP

    PTR 2F3A3

    COERCEDUP

    BINT2

    #<case

    DROPSWAPDROP

    BINT0

    3UNROLL

    ZERO_DO

    Z10_

    FPTR2 ^ZDIVext

    FPTR2 ^Z2BIN

    INDEX@

    BINT2

    #/

    DROP

    #0=

    IT

    ::

      #1+

      4PICKSWAP

      NTHCOMPDROP

    ;

    ROT#+

    BINT10

    #/

    DROP

    SWAPLOOP

    DROPSWAPDROP

    BINT10

    SWAP#-

    FPTR2 ^#>Z

  ;

;

Sorry, error in above programme, corrected here:

Code:

::

  CK1&Dispatch

  # FF

  ::

    DUP

    {

      BINT0

      BINT2

      BINT4

      BINT6

      BINT8

      BINT1

      BINT3

      BINT5

      BINT7

      BINT9

    }

    SWAPDUP

    ID x010

    DUP

    BINT2

    #<case

    DROPSWAPDROP

    BINT0

    3UNROLL

    ZERO_DO

    Z10_

    FPTR2 ^ZDIVext

    FPTR2 ^Z2BIN

    INDEX@

    BINT2

    #/

    DROP

    #0=

    IT

    ::

      #1+

      4PICKSWAP

      NTHCOMPDROP

    ;

    ROT#+

    BINT10

    #/

    DROP

    SWAPLOOP

    DROPSWAPDROP

    DUP#0=csedrp

    Z0_

    BINT10

    SWAP#-

    FPTR2 ^#>Z

  ;

;

Gerald H

Gerald H