IITB Practice Set: 26

Question

Consider a filesystem in which an inode of a file contains pointers to $12$ direct blocks, one single indirect block, and one double indirect block. The block size in the system is $64$ bytes, and a block number (i.e., a pointer to a block) is $4$ bytes in size. You may assume that the single and double indirect blocks store only block numbers (pointers) and no other metadata. What is the maximum size of a file (in blocks) that can be stored in this filesystem?

Answer 1

Take a look at the I-node structure here : Inode Structure

Let,

• Number of direct disk blocks be x. Given x = 12.
• Number of singly indirect blocks be y. Given y = 1
• Number of doubly indirect blocks be z. Given z = 1
• Number of pointers that can be stored in 1 block be P.

The maximum file size in number of blocks

                    = x + (y × P) + (z × $P^{2}$)

 

Comments

They are asking in blocks. What is the difference between these two ?

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I have updated the answer. To find in terms of number of blocks, we can divide the maximum file size/block size. We can directly find the number of blocks required for maximum file size by looking at the inode structure. For calculating the file size, we first find number of blocks and then multiply them with size of each block. Check the above updated answer.