santoshkumar

• Oct 23rd, 2014

 

the length is 8 byte it is find by using LENGTH OF clause.

Mona

• Nov 2nd, 2014

 

1-4=2 bytes
5-9=4 bytes
9 above 8 bytes
here 7+3=10+1(sign)
total 11 btes
so 8 bytes is the answer.

siddhu

• Nov 10th, 2014

 

yup for decimal point i.e.,each bit occupies half byte and for sing one extra half byte...so in total it occupies 10 halfbytes=5bytes...

r.sindhoor

• Dec 3rd, 2014

 

in my view comp clause doesn't support decimal values it supports only binary values.decimal values will be supported by comp-3

Rahul

• Jun 11th, 2015

 

I believe above question will occupy 8 bytes.
As for:
1 - 4 characters - 2 bytes
5 - 9 characters - 4 bytes
10 - 18 characters - 8 bytes
There is no space occupied by decimal.
The calculation which you are doing is for COMP -3 where in each character occupies half byte or a nibble

laxmanarao

• Jul 27th, 2015

 

6 Bytes

chinna

• Aug 12th, 2015

 

It occupies 10 bytes of space because no space allocated for decimal pointer v .

Riddhi

• Aug 12th, 2015

 

He looked at you differently since it was a trick question. The example he gave is invalid, since COMP only stores integers, no implied decimals.

aayush

• Feb 16th, 2016

 

8 byte because in comp 0-4 takes 2 bytes,5-9 takes 4 byte and 10-18 takes 8 byte.

marcelo

• Feb 18th, 2016

 

8 bytes.

MAGNO

• Aug 24th, 2016

 

There is no such length in cobol for comp. The max length is a full-word, which means S9(08) comp. 4 bytes long. For COMP-3 you may use 10 bytes long, and for Alpha 128 M

mani

• Aug 29th, 2016

 

PIC S9(01) TO S9(04) C0MP =2 BYTES
PIC S9(05) TO S9(09) C0MP =4 BYTES
PIC S9(10) TO S9(18) C0MP =8 BYTES

MUKESH

• Sep 1st, 2016

 

We cannot use pic clause when specifying comp usage clause...

Harika

• Feb 2nd, 2017

 

Comp is a binary storage format and does not allow decimals. It is invalid syntax

Ramya

• Feb 17th, 2017

 

Only numeric data is specified as COMP. I have 2 points to say here:
1) PIC X(3) COMP is not valid. Hence, this is invalid
2) Picture clause in COMP can have either 9 or S and here the picture clause being S9(3)V9(7) COMP becomes invalid

Vishwanath

• Feb 27th, 2017

 

8 Bytes..

Rajeshwar Reddy

• Jun 3rd, 2017

 

Hi,

It will take only 2bytes, because comp occupy only two bytes for 1 to 4 bytes.
9(3)v9(7) is 3.7bytes because V is a decimal point

Ganesh Dudhigani

• Jun 13th, 2017

 

It will take 8bytes for it.
9(1) to 9(4) = 2bytes
9(5) to 9(9) = 4bytes
9(10) to 9(18) = 8bytes

raghul kumar

• Aug 17th, 2017

 

As far as I learned comp cannot hold decimal values, if you give this statement compiler will show an error

ashlesha

• Aug 19th, 2017

 

COMP holds binary values and COMP has predetermined length to data types
9(1)-9(4) = 2 bytes
9(5)-9(9) = 4 bytes
9(10)-9(18) = 8 bytes..
9(3)v9(7) occupies 8 bytes but the question itself wrong COMP variable doesnt have a PIC clause bcz it already contains a predetermined data size.

Ganhi

• Mar 7th, 2018

 

Comp is integer values. Comp3 for decimal values.
decimal declaration with comp will give compiler error.

andyruni

• Aug 17th, 2018

 

Hello - Since there is a V , this computational usage must be COMP-3. Now in COMP-3 where it is signed or unsigned the sign is stored in the last 4 bits. 9(3)V9(7) COMP-3 will thus occupy 6bytes. Thank you.

GAURAVARAM

• Dec 5th, 2018

 

COMP will store the values in binary format and
from 9(1) -9(4) it will occupy 2 bytes
from 9(5) -9(8) it will occupy 4 bytes
from 9(9) -9(18) it will occupy 8 bytes