01 x pic s9(8) comp. How will the following value be internally allocated �

01 x pic s9(8) comp.
How will the following value be internally allocated
'18787'

Question Posted / velu1603

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01 x pic s9(8) comp. How will the following value be internally allocated '18787' ..

Answer / kapil arya

Basically Comp Makes the word boundary.
(0 - 4) 2 Bytes (Half Word boundary)
(5-9) 4 Bytes (Full Word Boundary)
(10-18) 8 byte Double word Boundary

Because it is s9(8) Comp..
So I will take 4 byte to store the date.
And it will allocated like:
01 87 87 , will store in 3 byte and one byte will be free
or unused..

Guys or Girls i have one question...
Related to JCL
I have a jcl where in a step i have one DD statement where
i am concatinating the 3 dataset..so in next step i want to
override the 3rd dataset only.. So how can i override the
3rd dataset.. Please let me know the answer...
Thanks to All..

Is This Answer Correct ?

9 Yes

0 No

01 x pic s9(8) comp. How will the following value be internally allocated '18787' ..

Answer / santhosh vayathuri

Guys or Girls i have one question...
Related to JCL
I have a jcl where in a step i have one DD statement where
i am concatinating the 3 dataset..so in next step i want to
override the 3rd dataset only.. So how can i override the
3rd dataset.. Please let me know the answer...
Thanks to All..

FOR ABOVE QUESTION ANSER IS

//S1 EXEC PGM=PROG1
//DD1 DD DSN = X.Y.Z
// DD DSN = A.B.C
// DD DSN = M.N.O

//S2 EXEC PGM= PROG2
//S1.DD1 DD
// DD
// DD DSN =SAN.THO.SH

PLEASE LET ME KNOW IF I AM WRONG

Is This Answer Correct ?

4 Yes

0 No

01 x pic s9(8) comp. How will the following value be internally allocated '18787' ..

Answer / krishna chaitanya

the answer one is correct . he asked a question , how to
override 3 rd step .

we can do lik ths

step3.ddname which will override the DS .

if any wrong correct me .

Is This Answer Correct ?

0 Yes

1 No

01 x pic s9(8) comp. How will the following value be internally allocated '18787' ..

Answer / sheeba

hey acording to me...its 4 bytes.... if we convert the
above value to binary format dats how it stores rite
so...2power4(1
)+2p3(8)+2p2(7)+2p1(8)+2p0(7)=131....so 3 bytes...and 1
unused... totally 4 bytes......

Is This Answer Correct ?

1 Yes

3 No

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