What does COMP VALUE ZERO in COBOL mean? - cobol

There is a great explanation for COMP under the following stackoverflow link:
https://stackoverflow.com/a/42423487/7802354
but I still don't understand what COMP VALUE ZERO in
77 ABC PIC S9(4) COMP VALUE ZERO.
means. I would appreciate if someone clarifies that.

COMP or COMPUTATIONAL refers to the storage representation. For most implementations, it is the same as (big-endian) BINARY, for some PACKED-DECIMAL, rarely it is the same
storage representation as DISPLAY. VALUE ZERO means that the initial value will be +0.
It is similar to
short abc = 0;
in some other languages. However the PIC S9(4) limits the value to -9999 through +9999.

The way to read that line of code
77 ABC PIC S9(4) COMP VALUE ZERO.
is
77 is the level
ABC is the data item name
PIC S9(4) COMP is the data item's 'data type'
VALUE ZERO is what to initialize the data item to.

Related

Store "Nº" before number in COBOL

I need to store Nº before a number (integer), like Nº423.
I try:
77 SC PIC 'Nº'9 VALUE 423.
77 SC PIC N9 VALUE 423.
But I don't achieve it. Can anyone help me?
01 prefixed-number.
05 the-prefix pic xx value 'Nº'.
05 the-number pic 999 value 423.
I suggest you familiarize yourself with your compiler's documentation on data definitions in COBOL, including the PICTURE clause and the USAGE clause.

COBOL supress last number while summing two decimal numbers

According to the COBOL code below when I try to sum WS-NUM1 with WS-NUM2, COBOL seems to supress the last number. For example: variable WS-NUM1 and WS-NUM2 are 10.15, I get 20.20 as result but expected 20.30. What's wrong?
WS-NUM1 PIC 9(2)V99.
WS-NUM2 PIC 9(2)V99.
WS-RESULTADO PIC 9(2)V99.
DISPLAY "Enter the first number:"
ACCEPT WS-NUM1.
DISPLAY "Enter the second number:"
ACCEPT WS-NUM2.
COMPUTE WS-RESULTADO = WS-NUM1 + WS-NUM2.
Thanks in advance.
PIC 9(2)v99 defines a variable with an implied decimal place not a real one. You're trying to enter data containing a decimal point and it's not working because you have to strip out the '.' to get the numeric part of your data to properly fit in the 4 bytes that your working storage area occupies.
PROGRAM-ID. ADD2.
data division.
working-storage section.
01 ws-num-input pic x(5).
01 WS-NUM1 PIC 9(2)V99 value 0.
01 redefines ws-num1.
05 ws-high-num pic 99.
05 ws-low-num pic 99.
01 WS-NUM2 PIC 9(2)V99 value 0.
01 redefines ws-num2.
05 ws-high-num2 pic 99.
05 ws-low-num2 pic 99.
01 WS-RESULTADO PIC 9(2)V99.
PROCEDURE DIVISION.
DISPLAY "Enter the first number:"
*
accept ws-num-input
unstring ws-num-input delimited by '.'
into ws-high-num, ws-low-num
DISPLAY "Enter the second number:"
accept ws-num-input
unstring ws-num-input delimited by '.'
into ws-high-num2, ws-low-num2
*
COMPUTE WS-RESULTADO = WS-NUM1 + WS-NUM2.
DISPLAY WS-RESULTADO
STOP RUN
.
This is just a simple demonstration. In a real world application you would have to insure much more robust edits to ensure that valid numeric data was entered.
If I declare it like this
01 WS-NUM1 PIC 9(2)V99.
01 WS-NUM2 PIC 9(2)V99.
01 WS-RESULTADO PIC 9(2)V99.
and define and sum them up like this
SET WS-NUM1 TO 10.15.
SET WS-NUM2 TO 10.15.
COMPUTE WS-RESULTADO = WS-NUM1 + WS-NUM2.
DISPLAY WS-RESULTADO.
I get the expected result of 20.30.
This looks like a job for a special type of PICture : Edited picture
Indeed you seem to know about the vanilla PICture clause (I'm writing PICture because as you may know it you can either write PIC or PICTURE).
A vanilla number PIC contains only 4 different symbols (and the parentheses and numbers in order to repeat some of the symbols)
9 : Represents a digit. You can repeat by using a number between parentheses like said before.
S : Means that the number is signed
V : Show the position of the implicit decimal point
P : I've been told that it exists but I honestly never found it in the codebase of my workplace. Its another kind of decimal point used for scaling factors but I don't know much about it.
But there are other symbols.
If you use theses other mysterious symbols the numeric PIC becomes an edited numeric PIC. As its name says, an edited PICture is made to be shown. It will allow you to format your numbers for better presentation or to receive number formatted for human reading.
Once edited, you cannot use it to make computations so you will have to transfer from edited to vanilla to perform computations on the latter. And you move from vanilla to edited in order to display your results.
So now I shall reveal some of these mysterious symbols :
B : Insert a blank at the place it is put
0 : Insert a zero at the place it is put
. : Insert the decimal point at the place it is put
: Insert a + if the number is positive and a - if the number is negative
Z : Acts like a 9 if the digits it represents has a value different than 0. Acts like a blank if the digits has the value of 0.
To my knowledge there are also : / , CR DB * $ -
You can look up for it on the internet. They really show the accountant essence of cobol.
For your problems we are really interested by the "." which will allow us to take into account the decimal point you have the write when you type down your input.
For a bonus I will also use Z which will make your result looks like 2.37 instead of 02.37 if the number is less than ten.
Note that you cannot use the repeating pattern with parenthesis ( 9(03) for instance) when describing an edited picture ! Each digits has to represented explicitly
IDENTIFICATION DIVISION.
PROGRAM-ID. EDITCOMP.
DATA DIVISION.
WORKING-STORAGE SECTION.
01 WS-NUM1-EDITED PIC 99.99.
01 WS-NUM2-EDITED PIC 99.99.
01 WS-NUM1-CALC PIC 9(2)V99.
01 WS-NUM2-CALC PIC 9(2)V99.
01 WS-RESULTADO-EDITED PIC Z9.99.
PROCEDURE DIVISION.
ACCEPT WS-NUM1-EDITED.
ACCEPT WS-NUM2-EDITED.
MOVE WS-NUM1-EDITED TO WS-NUM1-CALC.
MOVE WS-NUM2-EDITED TO WS-NUM2-CALC.
COMPUTE WS-RESULTADO-EDITED = WS-NUM1-CALC + WS-NUM2-CALC.
DISPLAY WS-RESULTADO-EDITED.
STOP RUN.
You should note that there also exist edited alphanumeric picture. You can insert Blank (B), zeroes (0) or / (/) in it.

How used DIVIDE with a CONDENSED NUMERIC Varibale in COBOL

My problem is: I have data in a database table. The column is: Z-ZYTL-RTPDHR defined as NOT NULL NUMBER(5,2) .
So I have a program that I need to move my data in one variable "H-ZYTL-RTPDHR" and after I will move this value divide by 100 in one column Z8 when i compile i obtained Excel with data in the different column.
My data:
Z8
------
34,28
70
97
8,57
21,43
94,28
94,28
100
40
40
what I should get:
Z8
-------
0,3428
0,7
0,97
0,0857
0,2143
0,9428
0,9428
0,100
0,40
0,40
my question how to declare the variable to obtain the good result?
Variables that I declared:
01 FILLER.
05 H-ZYTL-RTPDHR.
10 PIC S9(5)V9(2) comp-3 VALUE.
05 FILLER REDEFINES H-ZYTL-RTPDHR.
10 H-ZYTL-RTPDHR comp-3 pic s9(5)v99.
Equivalent Temps Plein
05 W-Z8 PIC -(5),99.
05 FILLER PIC X(001) VALUE ';'.
ALIM-WZ8 SECTION.
IF Z-ZYTL-NOMBRE > ZERO
IF TLCODTRA(Z-ZYTL-NOMBRE) NOT = SPACES
MOVE Z-ZYTL-RTPDHR(Z-ZYTL-NOMBRE) TO
H-ZYTL-RTPDHR
DISPLAY 'H-ZYTL-RTPDHR:' H-ZYTL-RTPDHR
DIVIDE H-ZYTL-RTPDHR BY 100 GIVING W-Z8
DISPLAY 'W-Z8 : ' W-Z8
END-IF
END-IF.
You have defined the name H-ZYTL-RTPDHR twice: that will confuse you and and the compiler as well.
Your second definition of H-ZYTL-RTPDHR is good for doing arithmetic. All you need is
01 FILLER.
10 H-ZYTL-RTPDHR comp-3 pic s9(5)v99.
Your first definition of H-ZYTL-RTPDHR defines a group level item, and you can't do arithmetic with it.
However, your question doesn't show the definition of Z-ZYTL-RTPDHR. If it's not numeric, then you won't get the results you expect.

Decode a Binary Coded Decimal

I have a field pic X(03) with a date in it as X'160101' format yymmdd.
I will like to know how to convert it to pic x(06).
So far I tried to move it back to a 9(03) comp and move the
9(03) comp to a 9(06) but it didn't work.
How can I do this?
What you have is a Binary Coded Decimal (BCD). That is, the data is held in a binary field, but it is the decimal representation, not the binary, which is important for the value. X'160101' in would be 1,442,049. X'160201' would be 1,442,305. So if you were able to treat it as a binary field (you'd have to prepend a binary zero to make it the correct length for a binary field) you'd have to do some silly calculation.
So you do something different, and easier. BCD is not a native COBOL datatype. PACKED-DECIMAL (often the same as COMP-3/COMPUTATIONAL-3) is a BCD-type which includes a sign-value in the low-order (right-most) half-byte. So not quite a BCD, but you can treat your BCD as PACKED-DECIMAL like this:
01 BCD-TO-PACKED.
05 BTP-SOURCE-BCD PIC XXX.
05 BTP-PACKED-ZERO PACKED-DECIMAL PIC 9 VALUE ZERO.
01 FILLER
REDEFINES BCD-TO-PACKED.
05 BTP-PACKED-TO-MOVE
PAKCED-DECIMAL PIC 9(6)V9.
01 DATE-AS-PIC-X PIC X(6).
01 DATA-AS-CHARACTER-NUMERIC
REDEFINES DATE-AS-PIC-X PIC 9(6).
MOVE your-source-value TO BTP-SOURCE-BCD
MOVE BTP-PACKED-TO-MOVE TO DATA-AS-CHARACTER-NUMERIC
After that you can happily reference DATE-AS-PIC-X to do whatever you want with it.
BCD-TO-PACKED is a four-byte group field which, through the VALUE clause and through nothing ever changing, has, in the last by X'0F'. That's an unsigned packed-decimal field with one digit and a value of zero.
Then you do the X-to-X MOVE for BTP-SOURCE-BCD. BCD-TO-PACKED now looks like this: X'1601010F'. Which is perfectly valid for a packed-decimal field. You could divide that by 10 to get your date (REDEFINES it as PACKED-DECIMAL PIC 9(7)) but why waste CPU?
Instead, in the REDEFINES you define one decimal place (V9). When the compiler generates the code to MOVE that to another numeric field with no decimal places, the source decimal place(s) are just dropped off. Presto! Divide by 10 without dividing.
Note: The data-names chosen are to aid the explanation. You should make yours meaningful to the data, not just use "trite" names. "Character numeric" is called USAGE DISPLAY in COBOL, it is the default if a USAGE is not supplied. The word USAGE is rarely used itself. So USAGE DISPLAY is what you have when there is a PIC X or PIC 9 field with no other usage.
Note: You should probably not be using two-digit years. You won't be able to deal with dates earlier than 2000 or later than 2199. Historical use of two-digit years was due to expensive disk storage and "data redundancy" (all century values were 19, so why hold the value 19 80 different times for the same account). If using two-digit years, ensure that there is something somewhere which is relevant to the century.
Note: If you attempt to MOVE a PIC X field to a numeric field, the compiler will assume that you have valid character numbers in the field, which is one reason your original attempt failed.
If you search stackoverflow, you should find the answer (this has already been answered).
But Create a fields like (my-date-x holds the date):
03 my-date-x pic x(3).
03 my-date-9 pic 9(6).
03 date-ymdv0 pic 9(6)v9 comp-3.
03 date-x pic x(3) redefines date-ymdv0.
And the code is
Move 0 to date-ymdv0
Move my-date-x to date-x
Move date-ymdv0 to my-date-9
The reason it works is for 9(6)v9 comp3 160101 is stored as x'1601010c'
which is what you have (+ a 0c at the end).

COBOL keyword REDEFINES

May I ask, what is the usage of the keyword REDEFINES in COBOL?
I can not understand the manual's definition.
What is the meaning of the following code?
01 WS_CHARGE_TXT_8X PIC X(08) VALUE "10000000".
01 WS_CHARGE_NUM_8 REDEFINES WS_CHARGE_TXT_8X.
05 WS_CHARGE_8 PIC 9(05)V9(03).
Thank you!
Basically Redefines reuses the spaces so in the above example WS_CHARGE_TXT_8X and WS_CHARGE_8 will point to the same block of memory. This allows you to look at a block of memory in different ways, in this case the variable can be viewed as a text PIC X and a signed numeric PIC S9. The -8X to -8 in the variable name is just stylistic to indicate the variable is being recast to another type or format to other programmers.
In the above example
the value of WS_CHARGE_TXT_8X is
"10000000"
the value of WS_CHARGE_8 is
of 10000.000.
If you moved 123.456 to WS_CHARGE_8 the value of WS_CHARGE_TXT_8X "00123456".
A more useful example is
03 Birth-Date-YYYYMMDD pic 9(8).
03 filler redefines Birth-Date-YYYYMMDD.
05 Birth-Date-YYYY pic 9(4).
05 Birth-Date-MM pic 99.
05 Birth-Date-DD pic 99.
In this case you can access the whole date Birth-Date-YYYYMMDD or the year / month / day individually (Birth-Date-YYYY etc).
It is analogous to union in 'C'.
It saves working storage space, and MOVE statements, and is also useful for transposing arrays of PIC(X), or accessing repeated fields as an array.
In the OP's case a numeric "type" is being created for the char contents of the prototype field.
A REDEFINES allows a different data description for storage which has already been defined.
That might be to allow entirely different data to be held there.
05 RL-AGENT-DATA.
...
05 RL-CUSTOMER-DATA REDEFINES RL-AGENT-DATA.
...
Or to give a description which allows a use of a part of the data, as in Bruce's example.
Or to use a piece of data that is the same, but for a different purpose:
05 INPUT-AMOUNT PIC X(10).
05 INPUT-AMOUNT-NUMERIC REDEFINES PIC 9(8)V99.
Whatever is in INPUT-AMOUNT it can be reported without problem, but only if it is a valid numeric (by testng it for NUMERIC), is INPUT-AMOUNT-NUMERIC used for some purpose.
Note that each 01 subsequent to the first under an FD is an "implicit REDEFINES".
Note also that items in the LINKAGE SECTION have the effect of "redefining" "something", even though if the address of the data is from a CALLing program, the definition is often the same as the original definition, and should usually at least match the PICtures of the original.

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