I ran into a strange statement when working on a COBOL program from $WORK.
We have a paragraph that is opening a cursor (from DB2), and the looping over it until it hits an EOT (in pseudo code):
... working storage ...
01 I PIC S9(9) COMP VALUE ZEROS.
01 WS-SUB PIC S9(4) COMP VALUE 0.
... code area ...
PARA-ONE.
PERFORM OPEN-CURSOR
PERFORM FETCH-CURSOR
PERFORM VARYING I FROM 1 BY 1 UNTIL SQLCODE = DB2EOT
do stuff here...
END-PERFORM
COMPUTE WS-SUB = I + 0
PERFORM CLOSE-CURSOR
... do another loop using WS-SUB ...
I'm wondering why that COMPUTE WS-SUB = I + 0
line is there. My understanding is that I
will always at least be 1
, because of the perform block above it (i.e., even if there is an EOT to start with, I
will be set to one on that initial iteration).
Is that COMPUTE
line even needed? Is it doing some implicit casting that I'm not aware of? Why would it be there? Why wouldn't you just MOVE I TO WS-SUB
?
Call it stupid, but with some compilers (with the correct options in effect), given
01 SIGNED-NUMBER PIC S99 COMP-5 VALUE -1.
01 UNSIGNED-NUMBER PIC 99 COMP-5.
...
MOVE SIGNED-NUMBER TO UNSIGNED-NUMBER
DISPLAY UNSIGNED-NUMBER
results in: 255. But...
COMPUTE UNSIGNED-NUMBER = SIGNED-NUMBER + ZERO
results in: 1 (unsigned)
So to answer your question, this could be classified as a technique used cast signed numbers into unsigned numbers. However, in the code example you gave it makes no sense at all.