FINDEND MACRO

Okay, macro time.   I have a string that can end with any of several different characters.

How do I find the length of the string, and the ending character.  AND, I might do this several places.
Oh, I know, I could write a macro.  OH, OH,  and include branch instructions to different places.
It uses both  SUBSTRING,   and  SUBLIST  notation.  SUBSTRING has the index outside the quotes,

and selects 1, 2, 3, ?? characters from within a list.
SUBLIST has the index inside the quotes, and selects 1 entry from a list of multiple entries.

In both cases,  &N  is the index.
The strings that I'm parsing, following LABEL=, separated by commas, look like:

LABEL=RECORD 120

LABEL=CARD 80,LINE 120,LGETMAIN 16

LABEL=CARD 12,LINE+12,CARD 8,LINE-8,CARD 8

LABEL=SAVECARD,GOTSY*,GETSY,GETSY+6

LABEL=GOTSYSIN,GOTLABEL*,DW 16

------------------------THIS IS WHAT'S IN THE SOURCE CODE. -----------------
F2 includes 'goto' labels.  F1  just returns an undex.

         LA    R14,DW         BEGINNING OF THE STRING

TESTF2   FINDEND  (14),16,PARAMS='+ ,-',                               X

               GOTO=(0,SAVEPM,0,GETNEXT-4,SAVEPM,0)

----------------------------------------------------------

TESTF1   FINDEND  (14),16,PARAMS='+ ,-' 

------------------------------------------------------

         LTR   R15,R1         Q. DID IT WORK?

         BZ    ERR               NO.

         SR    R15,R14           YES, CALC LENGTH

*                             WHEN DONE, REG-14 = BEG OF STRING

*                                        REG-1  = ENDING CHARACTER

*                                        REG-15 = LENGTH OF STRING

------------------- I put this in my program ----------------------------------

         $$LA  R14,card          BEGINNING OF THE STRING

TESTF2   FINDEND  (14),16,PARAMS='+ ,-',GOTO=(0,SAVEPM,0,GETNEXT-4,SAVEPM,0)

--------------------------This is what's generated--------------------

TESTF2   LA    1,0(14)

SYS0004A LA    1,1(1)

         LA    0,2*4

         CLI   0(1),C'+'

         BE    SAVEPM

         LA    0,3*4

         CLI   0(1),C' '

         BE    SYS0004X+2

         LA    0,4*4

         CLI   0(1),C','

         BE    GETNEXT-4

         LA    0,5*4

         CLI   0(1),C'-'

         BE    SAVEPM

         BCT   0,SYS0004A

SYS0004X SR    1,1

----------------------------------------------------

TESTF1    FINDEND  (14),16,PARAMS='+ ,-'  <== macro call (no GOTO=)

*-----------------------------------------------------------------

TESTF1   LA    1,0(14)       <====== generated

         LA    0,16

SYS0002A LA    1,1(1)

         CLI   0(1),C'+'

         BE    SYS0002X+2

         CLI   0(1),C' '

         BE    SYS0002X+2

         CLI   0(1),C','

         BE    SYS0002X+2

         CLI   0(1),C'-'

         BE    SYS0002X+2

         BCT   0,SYS0002A

SYS0002X SR    1,1

------------------------AND THEN I DID THIS FOR THIS SPECIFIC APPLICATION

         LTR   R15,R1     Q. DID IT WORK?

         BZ    ERR           NO.

         SR    R15,R14       YES, CALC LENGTH

         WHEN DONE, REG-14 = BEG OF STRING

                    REG-1  = END OF STRING

                    REG-15 = LENGTH OF STRING

-----------------------------but the MACRO does the magic--------------

         MACRO

&LBL     FINDEND  &FLD,&LEN,&REG=1,&PARAMS='+ ,-',&GOTO=

         LCLC  &L,&C,&G

&L       SETC  'SYS&SYSNDX'

         LCLA  &N,&M

&N       SETA  2

&M       SETA  K'&PARAMS

&LBL     $$LA  1,&FLD

&L.A     LA    1,1(1)

.LOOP    LA    0,&N*4

&C       SETC '&PARAMS'(&N,1)

&G       SETC  '&GOTO(&N)'

.*       MNOTE  ,' &N &M &PARAMS  &C  &G '

         CLI   0(1),C'&C'

         AIF   (T'&GOTO EQ 'O').BE 

         AIF   ('&GOTO(&N)' EQ '0').BE 

         BE    &G

         AGO   .NEX  

.BE      BE    &L.X+2

.NEX     ANOP 

&N       SETA  &N+1

         AIF  (&N LT &M).LOOP

         BCT   0,&L.A

&L.X     SR    1,1

         MEND

-----------------yes, I'm a happy camper.  I still got it.  :)---------