有人可以解释如何解决这个问题吗？
1. **What is the maximum size of a file that could be handled by single indirect and double indirect?**
Block size is 2K
Address size is 4 Bytes
12 slots for direct disk blocks
One slot each for single indirect and double indirect.
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2. **What is the maximum size of a file that could be handled by single indirect and double indirect?**
Block size is 4096 bytes can contain 1024 block references.
12 slots for direct disk blocks
One slot each for single indirect and double indirect.
我的尝试：
1.
Block size is 2K = 2**11
Address size is 4 Bytes = 2**2
in one block are (2**11/2**2)= 2**9 pointers
12 slots for direct disk blocks
One slot each for single indirect = 2**9
and double indirect. = 2**18
Maximum file size
( 12 + 2**9 + 2**18 ) * 2**11 = 2**29 ~ 536 MB
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2.
Block size is 4096 bytes can contain 1024 block references
12 slots for direct disk blocks
One slot each for single indirect = 1024 = 2**10
and double indirect. = 1024**2 = 2**20
Maximum file size
( 12 + 2**10 + 2**20 ) * 2**12 = 2**32
( 12 + 1024 + 1024**2 ) * 4096 = ~ 4GB
我仍然不确定这是否正确以及我错在哪里
我的资料来源是：
-https://stackoverflow.com/questions/2742163/maximum-file-size-given-a-pspecial-inode-struct
-https://cis.temple.edu/~ingargio/cis307/readings/stable.html
-https://www.cs.swarthmore.edu/~kwebb/cs45/s18/09-File_Systems.pdf