Answer / gopal

01 INPUT-REC.
02 REC-ROWS OCCURS 10 TIMES INDEXED BY I.
03 REC-COLS OCCURS 10 TIMES INDEXED BY J.
04 REC-DATA PIC X(20 ) - assume

PROCEDURE DIVISION.
READ FILE(INPUT) INTO(WORK-REC) - assume length as 100
SET I TO 1. SET J TO 1.
PERFORM 1000-WRITE 10 TIMES
STOP RUN.
1000-SECTION.
MOVE WORK-REC TO REC-DATA(I J).
SET J UP BY 1.
READ FILE(INPUT) INTO(WORK-REC).
This will move the 10 occurences of J that is (I 1) to (I
10)
Then the outer loop should be incremented 10 times that is
I from 1 to 10.

Answer / neilsh

It should be
01 xyz
05 pqr occurs 2 times indexed by i
07 abc occurs 50 times indexed by j
09 rec pic x(40).

so it would be like 50 occurance of rec and then 2 occurance
of abc

50 + 50 = two divisions

Answer / ameet

The answer for this is as below:

01 INPUT-REC.
02 REC-DETAIL OCCURS 100 TIMES INDEXED BY I.
03 REC-DATA PIC X(10 ) - assume

PROCEDURE DIVISION.
READ FILE(INPUT) INTO(WORK-REC) - assume length as 100
SET I TO 1.
PERFORM 1000-WRITE 100 TIMES
STOP RUN.
1000-SECTION.
MOVE WORK-REC TO REC-DATA(I).
SET I UP BY 1.
READ FILE(INPUT) INTO(WORK-REC).

i have the following varibles in the working storage 05 ws-A PIC X(30) VALUE 'ABCDEFGHIJKLMNOPQRESTUVWXYZ ' 05 WS-B REDEFINES WS-A 10 WS-B1 PIC X(10). 10 WS-B2 PIC 9(10). 10 WS-B3 PIC X(10). If I Display B1, B2 and B3 respectively, what is the value displayed in B2

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