Packed decimal fields can be converted to a character format using the ED instruction.  Additionally, editing symbols and features like commas, decimal points, and leading zero suppression can be included in the character version of the packed decimal field that is being edited.  The first step in editing a packed decimal field is to create an “edit word” which is a pattern of what the character output of the edit process should look like.  Typically, the edit word is moved to a field in the output buffer which is being built, prior to printing.  Then the packed decimal field is “edited” into the output field, destroying the copy of the edit word.


  First, we consider how to construct an appropriate edit word for a given packed decimal field.  This can be accomplished by defining a string of hexadecimal bytes that represents the edit word.  Each byte in the edit word corresponds to a byte in the edited character representation.  In creating the edit word there are a collection of standard symbols which are used to describe each byte:

          X’40’        This symbol, which represents a space, is usually coded as the first byte of the

                          edit word where it acts as a “fill character”.  The fill character is used to replace

                          leading zeroes which are not “significant”.


          X’20’        Called a “digit selector”, this byte represents a position in which a

                          significant digit from the packed field should be placed.


          X’21’        This hexadecimal byte represents a digit selector and a significance starter. 

                          Significance starts when the first non-zero digit is selected.  Alternatively, we

                          can force significance to start by coding a single x’21’ in the edit word.  In this

                          case, significance starts in the byte following the x’21’.  Significance is

                          important because every significant digit is printed, even if it is a leading zero.


          X’6B’        This is the EBCDIC code for a comma. 


          X’4B’        This is the EBCDIC code for a decimal point.


          X’60’        This is the EBCDIC code for a minus sign.  This symbol is sometimes coded on

                          the end of an edit word when editing signed fields.  The  x’60’ byte will be

                          replaced with the fill character if the number being edited is positive.  If the

                          number is in fact negative, the x’60’ will not be “filled”, and the negative sign will

                          appear in the edited output.


          X’C4C2’   The EBCDIC version of “DB” (Debit).  This functions like the x’60’.  Coding

                          these symbols at the end of an edit word causes “DB” to appear in the output

                          if the field being edited is negative, otherewise the “DB” is “filled”.


          X’C3D9’   The EBCDIC version of “CR” (Credit).  This functions like the Debit symbol

                          above.  When the number being edited is negative, the “CR” symbol will

                          appear in the edited output, otherwise it will be “filled”.

          X’5C’        The EBCDIC symbol for an asterisk.  This character is sometimes used as

                          a fill character when editing dollar amounts on checks.


   We now consider a sample edit word and the output it would create for several packed fields.


                    EDWD     DC    X’4020206B2021204B202060’

                    AOUT     DC    CL11

                    APK      DC    PL4’45387’  


Assume we execute the instructions below,


                    MVC      AOUT,EDWD

                    ED       AOUT,APK


First, the edit word is moved to a field in the output buffer.  Then the packed field is “edited” into the output field.   The results are illustrated in the diagram below.




The diagram indicates the results of the edit process:  The fill character (x’40’) is unaffected, and is left in its position.  The first decimal digit, 0, is “selected” by the first x’20’, and since leading 0’s are not significant, the x’20’ is replaced by the fill character.  The second digit, 0, is also selected, and it too, is filled with a x’40’.  Since significance has not started, the x’6B’ is filled with x’40’.  The first non-zero digit, 4, is selected and this signals that significance has started.  (Any non-zero digit which is selected turns on the significance indicator.)  Each digit after the 4 will appear in the edited result.  The “4” is replaced with its character equivalent - x’F4’.  The “5” is selected and its x’20’ is replaced with x’F5’.  The “3” is selected and is represented as x’F3’.  The x’4B’, a decimal point, remains unaffected.  The “8” is selected and is represented as x’F8’.  The “7” is selected and is represented as x’F7’.  Since the number being edited is positive, the x’60’ is filled with x’40’.   The final result would print as “    453.87 “. 



   Consider a second edit which uses the same edit word as in the previous example, but with a different value for APK.


                    EDWD     DC    X’4020206B2021204B202060’

                    AOUT     DC    CL11

                    APK      DC    PL4’-7’  


Again we execute the same sequence of instructions.


                    MVC      AOUT,EDWD

                    ED       AOUT,APK



As in every edit, the x’40’ fill character is unaffected by the edit process.  The first and second digits, both 0, are selected, and since they are leading 0’s and significance has not started, they are filled with x’40’.  The x’6B’ is also filled with x’40’ since significance has not started.  The next two digits, both 0, are selected and filled.  Since the x’21’ selected a leading 0, the significance indicator is turned on - significance starts with the next digit.  This means that all other digits will appear in a character representation, even if they are leading 0’s.  All other editing symbols will be printed as well.  The fifth digit, 0, is selected and represented as x’F0’.  The x’4B’ is preserved.  The next two digits, 0 and 7, are selected and represented as x’F0’ and x’F7’.  Finally, since the APK contains a negative number, the x’60’ is preserved.   The final result would print as “      0.07-”.



          Some Unrelated ED’s:


      APK       DC    PL2’123’   X’123C’

   AOUT      DS    CL4    

   AEDWD     DC    X’40202020’

                   ...             Result:

             MVC   AOUT,AEDWD

             ED    AOUT,APK       AOUT = X’40F1F2F3’ - ’ 123’


   BPK       DC    PL2’0’   X’000C’

   BOUT      DS    CL4    

   BEDWD     DC    X’40202020’

                   ...             Result:

             MVC   BOUT,BEDWD

             ED    BOUT,BPK       BOUT = X’40404040’ - ’    ’


   CPK       DC    PL2’0’   X’000C’

   COUT      DS    CL4    

   CEDWD     DC    X’40202120’

                   ...             Result:

             MVC   COUT,CEDWD

             ED    COUT,CPK       COUT = X’404040F0’ - ’  







   DPK       DC    PL2’0’   X’000C’

   DOUT      DS    CL4    




                   ...             Result:

             MVC   DOUT,DEDWD

             ED    DOUT,DPK       DOUT = X’5C5C5CF0’ - ’***0’



   EPK       DC    PL2’-30’   X’030D’  NEGATIVE NUMBER

   EOUT      DS    CL4    

   EEDWD     DC    X’40202120’

                   ...             Result:

             MVC   EOUT,EEDWD

             ED    EOUT,EPK       EOUT = X’4040F3F0’ - ’  30’



   FPK       DC    PL2’-30’   X’030D’   NEGATIVE NUMBER


   FEDWD     DC    X’4020212060’

                   ...             Result:

             MVC   FOUT,FEDWD

             ED    FOUT,FPK       FOUT = X’4040F3F060’ - ’  30-’


1.  There are two errors that beginners make when using ED :


     1)  The number of x’20’s and x’21’s does not match the number of decimal digits in the field

           being edited.  This is a critical error.  If the packed field has length “n”, the number of

           x’20’s and x’21’s is 2n - 1.  For example, if you are editing a packed field of length 6, the

           edit word must contain exactly 11 x’20’s and x’21’s.  A bad edit word will produce

           unpredictable output.


     2)  The output field size does not match the edit word size.  For example, suppose you coded

          the following,

                  AEDWD     DC   X’402020202120’

                  AOUT      DS   CL5

          When the edit word is moved to AOUT, the last byte of the edit word is not moved since

          the edit word is 6 bytes and the target field is 5 bytes.  The effect is that we are using an

         incorrect edit word, even though the definition of the edit word was correct.


2)   When editing, start with the packed field and design an edit word that matches it.  Then define the output field to match the edit word.  For example, if we start with a packed field of length 3 (5 decimal digits), we could design x’402021204B2020’ as an appropriate edit word (5 x’20’s and x’21’s).  Since the edit word is 7 bytes long, we would design a 7 byte output field to hold the edit word.

                   XPK       DS     PL3

                   XEDWD     DC     X’402021204B2020’

                   XOUT      DS     CL7


                             MVC    XOUT,XEDWD

                             ED     XOUT,XPK