问题
Suppose a UNIX file system has some constraints--say, 2 KB blocks and 8B disk addresses. What is the maximum file size if inodes contain 13 direct entries, and one single, double, and triple indirect entry each?
回答1:
This explains it for you:
http://www.cis.temple.edu/~ingargio/cis307/readings/stable.html
"The maximum size of a file will be 8KB*(10 + 2**10 + 2**20 + 2**30), that is more than 8TB."
Swap 8KB for your 2KB, and adjust the entries for the smaller block size.
2KB*(10 + 2**8 + 2**16 + 2**24)
It's not clear to me from your question if the 13 entries include the singles, doubles and triples, or if they are separate, but that should be easy to adjust -- just change the 10 in the expression to a 13.
I think I've adjusted all the math correctly... double check it =| Hope this isn't homework I did for you ;)
回答2:
How many pointers in 1 block?
each block is 2kb = 2^11
1 disk address is 8b = 2^3
So, in 1 block there are 2^11/2^3 = 2^8 pointers"
How many pointers in the file system?
for 13 direct entries = (2^8)*13 = 3328
for single = (2^8)^2 = 2^16
for double = (2^8)^3 = 2^24
for triple = (2^8)^4 = 2^32
total pointer is :3328 + 2^16 + 2^24 + 2^32"
Therefore the size of the file system is:
size of the disk is : total of pointer*size of the pointer , which is around 34 GB "
回答3:
We hv a shortcut that can be used for competitive exams, given as :
Max file size possible = [ block size/pointer size]^3 * block size
来源：https://stackoverflow.com/questions/2742163/maximum-file-size-given-a-particular-inode-structure