Question

A file system with 2 kilobyte blocks can access 32 gigabytes worth of data through an inode’s triple indirect block. How many bits does the file system use for block pointers?

Answer 1

File size = 32GB

32*2^30 bytes

=2^5*2^30

=2^35

Block size is =2KB

=2^11

so we know,

(Block size/Block address)^3 * Block size =Accessable file size

= (2^11/Block address)^3*2^11 = 2^35

= (2^11/Block Address)^3 = 2^24

= (2^11/Block Address) = 2^8

Block Address= 2^11/2^8

=2^3

=8

So, the bits used for block pointer is 8

Note-in formula power of 3 denote the inode triple indirect block