I am getting abend while comparing 2 fields
They are defined as:
01 A PIC S9(8) COMP VALUE 0.
01 B PIC S9(5) COMP-3.
The statement which is going into abend is:
PERFORM VARYING A FROM 1 BY 1 UNTIL A > B
THIS statement is in another loop and Value of B IS increased there. For first two iterations the value of B is correct. And for 3 rd iteration , The value of B Displayed is non numeric.
Could you please suggest what can be done here?
We have this kind of scenario where we need to compare 2 different numeric fields & sometimes non-numeric data creeps in from external systems. What we did is redefined the fields with an PIC X(n) equivalent to the byte size & kept a NUMERIC check on the IF statements. Some thing like this -
IF WS-FIELD1 NUMERIC AND WS-FIELD2 NUMERIC
IF WS-FIELD1 NOT = WS-FIELD2
*** both fields are numeric but not equal
ELSE
*** both fields are numeric & equal
END-IF
ELSE
IF WS-FIELD1-X NOT = WS-FIELD2-X
*** both fields are non-numeric & not equal
ELSE
*** both fields are non-numeric but equal
END-IF
END-IF
I'm not sure why you are using PERFORM VARYING for a numeric field but I'm sure you can fit it somewhere in the above code.
Related
I'm building a COBOL program to calculate the average of up to 15 integers. The execution displays a number that is far bigger than intended with a lot of trailing zeroes. Here is the relevant code:
Data Division.
Working-Storage Section.
01 WS-COUNTER PIC 9(10).
01 WS-INPUT-TOTAL PIC 9(10).
01 WS-NEXT-INPUT PIC X(8).
01 WS-CONVERTED-INPUT PIC 9(8).
01 WS-AVG PIC 9(8)V99.
Procedure Division.
PROG.
PERFORM INIT-PARA
PERFORM ADD-PARA UNTIL WS-COUNTER = 15 OR WS-NEXT-INPUT = 'q'
PERFORM AVG-PARA
PERFORM END-PARA.
INIT-PARA.
DISPLAY 'This program calculates the average of inputs.'.
MOVE ZERO TO WS-COUNTER
MOVE ZERO TO WS-INPUT-TOTAL
MOVE ZERO TO WS-AVG.
ADD-PARA.
DISPLAY 'Enter an integer or type q to quit: '
ACCEPT WS-NEXT-INPUT
IF WS-NEXT-INPUT NOT = 'q'
MOVE WS-NEXT-INPUT TO WS-CONVERTED-INPUT
ADD WS-CONVERTED-INPUT TO WS-INPUT-TOTAL
ADD 1 TO WS-COUNTER
END-IF.
AVG-PARA.
IF WS-COUNTER > 1
DIVIDE WS-INPUT-TOTAL BY WS-COUNTER GIVING WS-AVG
DISPLAY 'Your average is ' WS-AVG '.' WS-NEXT-INPUT
END-IF.
The reason I put WS-NEXT-INPUT as alphanumeric and move it to a numeric WS-CONVERTED-INPUT if the IF condition is satisfied is because I want it to be able to take "q" to break the UNTIL loop, but after the condition is satisfied, I want a numeric variable for the arithmetical statements. Here's what it looks like with the numbers 10 and 15 as inputs:
10is program calculates the average of inputs.
Enter an integer or type q to quit:
15
Enter an integer or type q to quit:
q
Your average is 1250000000.
The console is a bit buggy so it forces me to input the 10 in that top left corner most of the time. Don't worry about that.
You see my problem in that execution. The result is supposed to be 00000012.50 instead of 1250000000. I tried inserting a few of my other variables into that display statement and they're all basically as they should be except for WS-INPUT-TOTAL which with that combination of numbers ends up being 0025000000 instead of 0000000025 as I would have expected. Why are these digits being stored in such a weird and unexpected way?
You have that strange output because of undefined behavior - computing with spaces.
The MOVE you present has the exact same USAGE and same size - it will commonly be taken over "as is", it normally does not convert the trailing spaces by some magic, so WS-CONVERTED-INPUT ends up with 10 . As the standard says for the move:
De-editing takes place only when the sending operand is a numeric-edited data item and the receiving item is a numeric or a numeric-edited data item.
and if it would be an edited field then it still should raise an exception on the MOVE:
When a numeric-edited data item is the sending operand of a de-editing MOVE statement and the content of that data item is not a possible result for any editing operation in that data item, the result of the MOVE operation is undefined and an EC-DATA-INCOMPATIBLE exception condition is set to exist.
When computing with spaces you commonly would raise a fatal error, but it seems your compile does not have that activated (and because you didn't share your compile command or even your compiler, we can't help with that).
Different COBOL dialects often use (partial only when checks are not activated which would lead to an abort) zero for invalid data, at least for spaces (but they can use everything. This will then lead to WS-CONVERTED-INPUT "seen as" 10000000 - so your computation will then include those big numbers.
So your program should work if you enter the necessary amount of leading zeroes on input.
General:
"never trust input data - validate" (and error or convert as necessary)
at least if something looks suspicious - activate all runtime checks available, re-try.
Solution - Do an explicit conversion:
MOVE FUNCTION NUMVAL(WS-NEXT-INPUT) TO WS-CONVERTED-INPUT, this will strip surrounding spaces and then convert from left to right until invalid data is found. A good coder would also check up-front using FUNCTION TEST-NUMVAL, otherwise you compute with zero if someone enters "TWENTY".
I am learning some basics with open COBOL, and here is my question about PICTURE:
999V99 VALUE 12.4
displays "012.40"
I would also expect
99,999V99 VALUE 12.4 to produce some reasonable output, but the output is "00,01240" instead.
99,999V99 VALUE 1234.56
also displays "01,23456" and not "01,234.56" as I would expect.
What is wrong? What is the correct mask to obtain "01,234.56" from 1234.56?
The correct mask is 99,999.99. Have a look at the link Gilbert suggested in the comments
Common mask characters
9 - 0 -> 9
Z - 0 -> 9 + space for leading zero's
V - Assumed decimal place not actually represented
. - actual decimal place
, - comma
So
Cobol definition Display value
---------------- -------------
99,999V99 VALUE 1234.56 01,23456
99,999.99 VALUE 1234.56 01,234.56
ZZ,ZZ9.99 VALUE 1234.56 1,234.56
--,--9.99 VALUE 1234.56 1,234.56
--,--9.99 VALUE -1234.56 -1,234.56
++,++9.99 VALUE 1234.56 +1,234.56
Once you use characters like Z , . - + / the field is and edited numeric field rather than a numeric field. Some compilers may let you use edited numeric in numeric calculations other will not. edited numeric are for displaying values.
Numeric fields
Numeric fields are defined like the following and can be used in numeric calculations
03 num-1 pic s9(5)v99.
03 num-2 pic s9(5)v99 comp.
03 num-3 pic s9(5)v99 comp-3.
I realize I have asked a similar question before, but the whole thing is more complicated than I thought.
To cut to the chase, I need to convert a string that contains numbers and letters into a string that only contains numbers, while keeping the numbers that were already there, in the right position.
The letters need to be converted to their corresponding position in the Alphabet + 9. So, A = 10, B= 11.... Z = 35.
So, basically, a string that looks like this:
'GB00LOYD1023456789A1B2'
will have to become:
'161100212429131023456789101112'.
I bolded the letters in both examples so you can see the difference more clearly. Depending on the input, the content will be longer or shorter than this example. Letters will be alternated by numbers and vice versa.
What's the best way to do this?
What's the best way to do this?
That is a matter of opinion.
The REPLACING option of the INSPECT verb requires the replacing and replaced character strings to be the same size, so that's right out because you need to replace one character with two. This is true at least for IBM COBOL.
A way to do this would be to loop through your input string and do a class check on each character. Something like...
01 Stuff.
05 in-posn pic s999 packed-decimal value +0.
05 out-posn pic s999 packed-decimal value +1.
05 in-string pic x(022) value 'GB00LOYD1023456789A1B2'.
05 out-string pic x(100) value spaces.
05 replacer pic x(002) value spaces.
perform varying in-posn from 1 by 1
until in-posn > length of in-string
if in-string(in-posn:1) alphabetic
evaluate in-string(in-posn:1)
when 'A' move '10' to replacer
when 'B' move '11' to replacer
.
.
.
when 'Z' move '35' to replacer
end-evaluate
string replacer delimited size
into out-string
pointer out-posn
end-string
else
string in-string(in-posn:1) delimited size
into out-string
pointer out-posn
end-string
end-if
end-perform
There are variations available. You could replace the evaluate with a couple of table lookups. You could store the length of in-string before beginning the loop. You could store in-string(in-posn:1) rather than hoping the compiler will do that for you.
This is just freehand but I think it conveys the idea.
I have the following variable in COBOL program which gets its value from a file, when it is read:
01 Employee-number PIC 09(8) comp
01 Employee-number-x redefines
Employee-number PIC x(04)
I have another variable in the same program:
01 D-element-number PIC 9(04)
Now,
Move Employee-number to D-element-number
Then I write this D-element-number to a file
value read from input file is :
0013
0024
so this value comes to Employee-number and Employee-number-x and I move this value to D-Element-number and write this variable to a output file.
But I get this in the output file:
4660
FFFF
4660
4660 is the decimal equivalent of X'1234'
But I want to see something like:
1234
FFFF
1234
How can I achieve this ?
I am allowed to change the definition of D-element-number but nothing else.
Assuming that when you have X'00001234' you want C'00001234', have a look here. http://ibmmainframes.com/viewtopic.php?p=234231#234231
Ignore the rest of the discussion for now, just concentrate on that post.
This is the key part:
PERFORM UNTIL X-WS-1000 > X-WS-1000-MAX
MOVE WS-1000-BYTE-TBL (X-WS-1000)
TO WS-PACKED-X (1:1)
MOVE WS-PACKED TO WS-DISPLAY
MOVE WS-DISPLAY-X TO WS-2000-BYTE-TBL (X-WS-2000)
SET X-WS-1000 UP BY 1
SET X-WS-2000 UP BY 1
END-PERFORM
You need the exact storage definitions to go with it (don't "correct" anything).
It works by getting the compiler to use the "unpack" instruction (UNPK). This one is working one byte at a time, so simple to explain.
X'12' (an example on-byte field) is put in a two-byte field. X'12ii' (where the value of ii is "irrelevant").
UNPK will then turn this "packed" number into a "zoned" number. A "Zone" is the first four bts of a byte, and for a Zoned number, the four bits are all set to one. So you get an F. Then you get the left-most digit from the first byte. Then you get the second output byte, first four bits set to F, then the second input digit.
Then the UNPK continues with the final byte, which contains one irrelevant digit, and an irrelevant sign. For a Zoned number, the sign and right-most digit occupy the same byte (the sign in the zone) so you get a whole byte of irrelevance.
X'12' -> X'12ii' -> X'F1F1ii'.
The first two bytes of the three-byte output are C'12'.
Nos, what is all fine for numbers, but letters make a mess:
X'AB' -> X'ABii' -> X'FAFBii'
Although F and a digits gives a displayable number, for F and a letter, the result does not mean much directly.
Now the INSPECT ... CONVERTING comes to the rescue: FA gets translated to C'A' (X'C1), and the same for the letters through to F.
Your results after the CONVERTING will be C'AB'.
Should give you enough to work on.
There are other methods, but this is a fair COBOL approximation to the classic Assembler technique with UNPK and a TRANSLATE (TR) and a table of translation values.
If you use your favourite search engine. you should be able to fine more methods, using calculation (more than one), table-lookups, I've even seen a 256-WHEN EVALUATE that "works", but I guess it is a little on the "slow" side.
Thinking further, you actually have a BCD (Binary Coded Decimal) don't you? This is a Packed Decimal, without the sign. You don't have any alphas in your field.
This is even simpler to convert.
01 the-converted-value PACKED-DECIMAL PIC 9(8)V9 VALUE ZERO.
01 FILLER REDEFINES the-converted-value.
05 the-binary-value PIC X(4).
05 FILLER PIC X.
MOVE Employee-number-x TO the-binary-value
MOVE the-converted-value TO D-element-number (with Gilbert's correction to PIC 9(8)).
The "decimal place" is the ii, the ignore value and ignore sign. You no longer need the INSPECT ... CONVERTING ... as you only have numeric digits. If you have a BCD...
A really good way for you to have answered your own question would have been to find out how the number was created in the binary field in the first place.
Didn't try this, but I think it might work:
01 WN-GROUP.
05 Employee-number PIC 9(8) comp.
05 PACKED-ZERO PIC 9(1) COMP-3 VALUE ZERO.
01 WN-PACKED redefines WN-GROUP PIC 9(9) COMP-3.
01 WN-UNPACKED PIC 9(9).
Then in your procedure:
MOVE your-number TO Employee-number
MOVE WN-PACKED TO WN-UNPACKED.
And finally you can move or display WN-PACKED(1:8).
When I move a number in a PIC X to a PIC 9 the numeric field's value is 0.
FOO, a PIC X(400), has '1' in the first byte and spaces in the remaining 399. Moving into the PIC 9(02) BAR like so
DISPLAY FOO
MOVE FOO to BAR
DISPLAY BAR
yields
1
0
Why is BAR 0 instead of 1? [Edit: originally, 'What is happening?']
Postscript: NealB says "Do not write programs that rely on obscure truncation rules and/or
data type coercion. Be precise and explicit in what you are doing."
That made me realize I really want COMPUTE BAR AS FUNCTION NUMVAL(FOO) wrapped in a NUMERIC test, not a MOVE.
Data MOVEment in COBOL is a complex subject - but here is
a simplified answer to your question. Some data movement rules
are straight forward and conform to what one might expect. Others are somewhat bizzar and may vary with
compiler option, vendor and possibly among editions of the COBOL standard (74, 85, 2002).
With the above in mind, here is an explanation of what happend in your example.
When something 'large' is
MOVEd into something 'small' truncation must occur. This is what happened when BAR was MOVEd to FOO. How that
truncation occurs is determined by the receving item
data type. When the receiving item is character data (PIC X), the rightmost characters will be truncated from the sending field.
For numeric data the leftmost digits are truncated from the sending field. This behaviour is pretty much universal for all COBOL
compilers.
As a consequense of these rules:
When a long 'X' field (BAR) starting with a '1' followed by a bunch of space characters is MOVEd
into a shorter 'X' field the leftmost characters are transferred. This is why the '1' would be preserved when moving to another PIC X
item.
When a long 'X' field (BAR) is moved to a '9' (numeric) datatype the rightmost characters are moved first. This is why '1' was lost, it was never
moved, the last two spaces in BAR were.
So far simple enough... The next bit is more complicated. Exactly what happens is vendor, version, compiler option and character set
specific. For the remainder of this example I will assume EBCDIC character sets and the IBM Enterprise COBOL compiler are being used. I
also assume your program displayed b0 and not 0b.
It is universally legal in COBOL to move PIC X data to PIC 9 fields provided the PIC X field contains only digits. Most
COBOL compilers only look at the lower 4 bits of a PIC 9 field when determining its numeric value. An exception is the least
significant digit where the sign, or lack of one, is stored. For unsigned numerics the upper 4 bits of the least significant digit
are set to 1's (hex F) as a result of the MOVE (coercion follows different rules for signed fields). The lower 4 bits are MOVEd without
coercion. So, what happens when a space character is moved into a PIC 9 field? The hex
representation of a SPACE is '40' (ebcdic). The upper 4 bits, '4', are flipped to 'F' and the lower 4 bits are moved as they are. This results in the
least significant digit (lsd) containing 'F0' hex. This just happens to be the unsigned numeric representation for the digit '0' in a PIC 9 data item.
The remaining leading digits are moved as they are (ie. '40' hex). The net result is that FOO displays as
b0. However, if you were to do anything other that 'MOVE' or 'DISPLAY' FOO, the upper 4 bits of the remaining 'digits' may be coerced to zeroes as a
result. This would flip their display characteristics from spaces to zeros.
The following example COBOL program and its output illustrates these points.
IDENTIFICATION DIVISION.
PROGRAM-ID. EXAMPLE.
DATA DIVISION.
WORKING-STORAGE SECTION.
01.
05 BAR PIC X(10).
05 FOO PIC 9(2).
05 FOOX PIC X(2).
PROCEDURE DIVISION.
MOVE '1 ' TO BAR
MOVE BAR TO FOO
MOVE BAR TO FOOX
DISPLAY 'FOO : >' FOO '< Leftmost trunctaion + lsd coercion'
DISPLAY 'FOOX: >' FOOX '< Righmost truncation'
ADD ZERO TO FOO
DISPLAY 'FOO : >' FOO '< full numeric coercion'
GOBACK
.
Output:
FOO : > 0< Leftmost trunctaion, lsd coercion
FOOX: >1 < Righmost truncation
FOO : >00< full numeric coercion
Final words... Best not to have to know anything about this sort to thing. Do not write programs that rely on obscure truncation
rules and/or data type coercion. Be precise and explicit in what you are doing.
Firstly, why do you think it might be useful to MOVE a 400-byte field to a two-byte field? You are going to get a "certain amount(!)" of "truncation" with that (and the amount of truncation is certain, at 398 bytes). Do you know which part of your 400 bytes is going to be truncated? I'd guess not.
For an alpha-numeric "sending" item (what you have), the (maximum) number of bytes used is the maximum number of bytes in a numeric field (18/31 depending on compiler/compiler option). Those bytes are taken from the right of the alpha-numeric field.
You have, therefore, MOVEd the rightmost 18/31 digits to the two-digit receiving field. You have already explained that you have "1" and 399 spaces, so you have MOVEd 18/31 spaces to your two-digit numeric field.
Your numeric field is "unsigned" (PIC 9(2) not PIC S9(2) or with a SIGN SEPARATE). For an unsigned field (which is a field with "no operational sign") a COBOL compiler should generate code to ensure that the field contains no sign.
This code will turn the right-most space in your PIC 9(2) into a "0" because and ASCII space is X'20' and an EBCDIC space is X'40'. The "sign" is embedded in the right-most byte of a USAGE DISPLAY numeric field, and and no other data but the sign is changed during the MOVE. The 2 or 4 of X'2n' or X'4n' is, without regard to its value, obliterated to the bit-pattern for an "unsign" (the lack of an "operational sign"). An "unsign" followed by a numeric digit (which is the '0' left over from the space) will, obviously, appear as a zero.
Now, you show a single "1" for your 400-byte field and a single 0 for your two-byte numeric.
What I do is this:
DISPLAY
">"
the-first-field-name
"<"
">"
the-second-field-name
"<"
...
or
DISPLAY
">"
the-first-field-name
"<"
DISPLAY
">"
the-second-field-name
"<"
...
If you had done that, you should find 1 followed by 399 spaces for your first field (as you would expect) and space followed by zero for your second field, which you didn't expect.
If you want to specifically see this in operation:
FOO PIC X(400) JUST RIGHT.
MOVE "1" TO FOO
MOVE FOO TO BAR
DISPLAY
">"
FOO
"<"
DISPLAY
">"
BAR
"<"
And you should see what you "almost" expect. You probably want the leading zero as well (the level-number 05 is an example, whatever level-number you are using will work).
05 BAR PIC 99.
05 FILLER REDEFINES BAR.
10 BAR-FIRST-BYTE PIC X.
88 BAR-FIRST-BYTE-SPACE VALUE SPACE.
10 FILLER PIC X.
...
IF BAR-FIRST-BYTE-SPACE
MOVE ZERO TO BAR-FIRST-BYTE
END-IF
Depending on your compiler and how close it is to ANSI Standard (and which ANSI Standard) your results may differ (if so, try to get a better compiler), but:
Don't MOVE alpha-numeric which are longer than the maximum a numeric can be to a numeric;
Note that in the MOVE alpha-numeric to numeric it is the right-most bytes of the alpha-numeric which are actually moved first;
An "unsigned" numeric should/must always remain unsigned;
Always check for compiler diagnostics and correct the code so that no diagnostics are produced (where possible);
When showing examples, it is highly important to show the actual results the computer produced, not the results as interpreted by a human. " 0" is not the same as "0 " is not the same as "0".
EDIT: Looking at TS's other questions, I think Enterprise COBOL is a safe bet. This message would have been issued by the compiler:
IGYPG3112-W Alphanumeric or national sending field "FOO" exceeded 18 digits. The rightmost 18 characters were used as the sender.
Note, the "18 digits" would have been "31 digits" with compiler option ARITH(EXTEND).
Even though it is a lowly "W" which only gives a Return Code of 4, not bothering to read it is not good practice, and if you had read it you'd not have needed to ask the question - although perhaps you'd still not know how you ended up with " 0", but that is another thing.
I gather you expect the 9(2) value to show up as "1" instead of "0" and you are confused as to why it does not?
You are moving values from left to right when you move from an X value (unless the destination value changes things). So the 9 value has a space in it. To simplify it, moving "X(2) value '1 '" to a 9(2) value literally moves those characters. The space makes what is in the 9(2) invalid, so the COBOL compiler does with it what it knows to do, return 0. In other words, defining the 9(2) as it does tells the compiler to interpret the data in a different way.
If you want the 9(2) to show up as "1", you have to present the data in the right way to the 9(2). A 9(2) with a value of 1 has the characters "01". Untested:
03 FOO PIC X(2) value '1'.
03 TEXT-01 PIC X(2) JUSTIFIED RIGHT.
03 NUMB-01 REDEFINES TEXT-01 PIC 9(2).
03 BAR PIC 9(2).
DISPLAY FOO.
MOVE FOO TO TEXT-01.
INSPECT TEXT-01 REPLACING LEADING ' ' BY '0'.
MOVE NUMB-01 TO BAR.
DISPLAY BAR.
Using the NUMERIC test against BAR in your example should fail as well...