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Suppose i want to declare a binary comp fild of 7 byte .how
to write?
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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.
| Is This Answer Correct ? | 14 Yes | 0 No |
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.
| Is This Answer Correct ? | 4 Yes | 0 No |
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.
| Is This Answer Correct ? | 6 Yes | 6 No |
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
| Is This Answer Correct ? | 1 Yes | 2 No |
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