MadeEasy Subject Test 2019: Operating System - File System

Question

Unix I-node has block size 8 KB and file possible with triple indirect is 128 GB.Number of bits disk block address contain is?

Comments

yes correct 256 bits.

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Can u give sol. Plz

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@pawan kumarln, to address one block no. of pointers needed = disk block size/ disk block address(p)

as maximum possible file size is given in question with triple indirect so,

$(\frac{8KB}{P})^{3}*8KB(block size) = 128GB$

solve it you'll get p = 256 bits.

Answer 1

See here triple indirect is given,

For triple indirect max possible file size is

= {(block size) /DBA}^3 * Block size

// DBA:Disk Block Address

So, 128 GB= (8 KB/DBA) ^3 *8KB

(DBA)^3 = 2^15 B

Cube root both side, and you will get

DBA =32 Bytes = 32*8 bits=256 bits