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Memory size allocation for Comp data

 
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prabhucs01
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Joined: 07 Mar 2005
Posts: 43

PostPosted: Sun Jan 13, 2008 1:23 pm    Post subject: Memory size allocation for Comp data
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Number of 9's length in bytes

1 to 4 2 Bytes(halfword)
5 to 9 4 Bytes(fullword)
10 to 18 8 Bytes(double word)

if the first four digits require 2 bytes(4bits/digit) then for total 9 digits it must be 36 bits and for 18 must be 72 bits.

could anyone give a detail about hw its been stored. i have checked with the below link too

http://ibmmainframes.com/size.html

sorry if the question is been asked before too. please help me in knowing the way the digits are stored.
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Craq Giegerich

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Joined: 19 May 2007
Posts: 1512
Location: Virginia, USA

PostPosted: Sun Jan 13, 2008 6:57 pm    Post subject: Re: Memory size allocation for Comp data
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prabhucs01 wrote:
Number of 9's length in bytes

1 to 4 2 Bytes(halfword)
5 to 9 4 Bytes(fullword)
10 to 18 8 Bytes(double word)

if the first four digits require 2 bytes(4bits/digit) then for total 9 digits it must be 36 bits and for 18 must be 72 bits.

could anyone give a detail about hw its been stored. i have checked with the below link too

http://ibmmainframes.com/size.html

sorry if the question is been asked before too. please help me in knowing the way the digits are stored.


There is not a direct correspondence between the number of decimal digits and the number of binary bits. 8 bits (1 byte) can hold some three decimal digit numbers, 16 bits (2 bytes) can hold some 5 decimal digit numbers.
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