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prabhucs01 Warnings : 1 New User
Joined: 07 Mar 2005 Posts: 43




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




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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