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ravi.veda
New User
Joined: 30 Mar 2007
Posts: 7
Location: banglore
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01 AMOUNT S9(12)V99
AMOUNT is the variable, if it is positive fine. If it is negative, the sign will be store in the last i.e rightmost side.
ex: i/p '-12345.67'
the o/p should be '12345.67-' |
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stodolas
Active Member
Joined: 13 Jun 2007
Posts: 632
Location: Wisconsin
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You need to switch to display format
01 Amount PIC -9(12).99. |
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dick scherrer
Moderator Emeritus
Joined: 23 Nov 2006
Posts: 19244
Location: Inside the Matrix
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Hello,
I'd most likely use
01 OUTPUT-VALUE PIC Z(12).99- to suppress the leading zeros and place the minus at the right end.
This "01 AMOUNT S9(12)V99 " will not contain "i/p '-12345.67' ". The "AMOUNT" field can only contain numerics, not a "-". The value might be negative, but value will not contain a "-". |
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ravi.veda
New User
Joined: 30 Mar 2007
Posts: 7
Location: banglore
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hi scherrer,
thank you,iam getting the output by using
01 OUTPUT-VALUE PIC Z(12).99-
Can you plz guide me wat's the wrong in given below.
i have variable1 with s9(12)v99 and variable2
if var1<0
move var1(1:1) to var2(16:1)
move var1(2:13) to var2(1:12)
move var1(15:2) to var2(14:2)
ex:var1=-666661234566.33 |
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Marso
REXX Moderator
Joined: 13 Mar 2006
Posts: 1353
Location: Israel
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PIC S9(12)V99 does not mean there is a sign in the leftmost character,
it means the number is signed. The sign is kept in the rightmost character.
So var1 contains: '6666612345663L' and you're just making a big mess with the number  |
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Craq Giegerich
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Joined: 19 May 2007
Posts: 1512
Location: Virginia, USA
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ravi.veda,
The size of a s9(12)v99 is 14 bytes not 16, the sign does not occupy a byte and the V implies a decimal point that does not exist. Have you considered looking in the COBOL manuals? |
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dick scherrer
Moderator Emeritus
Joined: 23 Nov 2006
Posts: 19244
Location: Inside the Matrix
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Hello,
| Quote: |
if var1<0
move var1(1:1) to var2(16:1)
move var1(2:13) to var2(1:12)
move var1(15:2) to var2(14:2)
ex:var1=-666661234566.33 |
If you are going to use variables as signed numbers in some calculation, you will most likely not use reference modification to re-arrange the data.
If you more completely explain what you have and what you want to do with it, we can probably give better replies. I suspect there is more than what you have posted so far. What else needs to happen with the s(12)v99 field content?
| Quote: |
thank you,iam getting the output by using
01 OUTPUT-VALUE PIC Z(12).99- |
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ravi.veda
New User
Joined: 30 Mar 2007
Posts: 7
Location: banglore
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Hischerrer,
y am asked is,before ur answer i think like that.so,i expressed my question and get clear clarification.
actually it is a part of my CR.
 Really Thank You all  |
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dick scherrer
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Joined: 23 Nov 2006
Posts: 19244
Location: Inside the Matrix
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| You're welcome, good luck |
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Mickeydusaor
Active User
Joined: 24 May 2006
Posts: 258
Location: Salem, Oregon
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Have you tried
input s9(12)v99 sign is leading
output s9(12)v99 sign is trailing
move input to output |
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annujp
New User
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Joined: 31 Aug 2005
Posts: 39
Location: St Paul,MN
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Just thought i wud try this out.
I declared these variables.
| Code: |
01 WS-A PIC S9(4)V99 VALUE ZEROS
SIGN LEADING SEPARATE.
01 WS-B PIC S9(4)V99 VALUE ZEROS
SIGN TRAILING SEPARATE.
01 WS-C PIC -9(4).99 VALUE ZEROS.
01 WS-D PIC 9(4).99- VALUE ZEROS. |
I moved the variables to switch signs.
| Code: |
MOVE -1234.99 TO WS-A
WS-C.
MOVE WS-A TO WS-B.
MOVE WS-C TO WS-D.
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This was my result
| Code: |
WS-A -123499
WS-B 123499-
WS-C -1234.99
WS-D 1234.99- |
Sign leading and sign trailing separate will occupy another byte for the sign. So S9(4)V99 SIGN LEADING SEPARATE will occupy 7 bytes, with the sign in the first byte.
Correct me if I am wrong. |
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jmreddymca
Warnings : 1
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Joined: 14 Oct 2007
Posts: 29
Location: Bangalore
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| yes Anitha you are right , we can do the way you suggested |
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HARLEEN SINGH MANN
Warnings : 2
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Joined: 03 Aug 2007
Posts: 17
Location: Pune
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| ravi.veda wrote: |
01 AMOUNT S9(12)V99
AMOUNT is the variable, if it is positive fine. If it is negative, the sign will be store in the last i.e rightmost side.
ex: i/p '-12345.67'
the o/p should be '12345.67-' |
There is one mistake the writere has made here, the -ve sign is not stored after 7 but in the same byte where 7 is. so the display wont be 12345.67- but 123456* (some char in place of *). This is due to the defined type of AMOUNT.
To achieve the op you require define it or move it into a
S9(12)V99 SIGN TRAILING SEPERATE.
Hope things are clearer now.
Regards. |
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