Answer / i go crazy

Hi Suv,

If u code S9(14) COMP, it will take double word i.e. 8
bytes. Below is the size of comp variables.
S9(01) - S9(04) : 2 BYTES.
S9(05) - S9(09) : 4 BYTES.
S9(10) - S9(18) : 8 BYTES.

Answer / tata

if we use comp3(packed decimal)
declare like 01 a pic s9(13)comp3.
13/2=6.5 and s takes extrahalf byte.
that is 6.5+0.5=7.

Answer / suv

If your question is like this:::
Suppose i want to declare a binary COMP field of storage 7
bytes .how
to write?


Answer is:
PIC S9(14) COMP.

storage byte required by COMP field will = half of the
digit specified in the PIC clause.
Just an info....
storage byte required by COMP3 field will = ( half of the
digit specified in the PIC clause)+1

Please correct me if I am wrong.

Answer / ankit

hi i think you cannot declare binary comp of 7 bytes
exactly as the comp variable are fixed to half(2 bytes)
word,full word(4 bytes) and double word(8 bytes), so you
need to choose the paricular comp variable to use ok
correct me if i am wrong

Answer / sudheer

COmp will take 2byte pic size-1-4
4byte pic size 5-9
8byte pic size 10-18

if we use comp-3 it will take n/2+1/2 byte extra for sign if
pic is odd no.if it is even it will take n/2.

my answer is 8 bytes

pls let me know if any thng wrong

Answer / sky

you would declare it as s9(7) comp

01 a pic 9(9v99) 01 b pic 9(9.99) wht will be the stored vales in both cases

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