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While reading this, I found a reasonable answer, which says:

Case 1: Directly Writing to File On Disk

100 times x 1 ms = 100 ms

I understood that. Next,

Case 3: Buffering in Memory before Writing to File on Disk

(100 times x 0.5 ms) + 1 ms = 51 ms

I didn't understand the 1 ms. What is the difference in between writing 100 data to disk and writing 1 data to disk? Why do both of them cost 1 ms?

Community
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cengizhan
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    The difference is in actually accessing the place where you write the data. An analogy would be writing data would require you to walk nextdoor and write on a black board. If I tell you to go write "foo" on there, wait until you go there, come back, then ask you to write "bar", wait again, repeat that 100 times; It's faster to give you a list of things to write there and come back when you're done. You save 99 times going back and forth – Eregrith Apr 27 '15 at 09:26

4 Answers4

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The disc access (transferring data to disk) does not happen byte-by-byte, it happens in blocks. So, we cannot conclude if that the time taken for writing 1 byte of data is 1 ms, then x bytes of data will take x ms. It is not a linear relation.

The amount of data written to the disk at a time depends on block size. For example, if a disc access cost you 1ms, and the block size is 512 bytes, then a write of size between 1 to 512 bytes will cost you same, 1 ms only.

So, coming back to the eqation, if you have , say 16 bytes of data to be written in each opeartion for 20 iteration, then,

  • for direct write case

time = (20 iteration * 1 ms) == 20 ms.

  • for buffered access

time = (20 iteration * 0.5 ms (bufferring time)) + 1 ms (to write all at once) = 10 + 1 == 11 ms.

Sourav Ghosh
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It is because of how the disc physical works. They can take larger buffers (called pages) and save them in one go. If you want to save the data all the time you need multiple alteration of one page, if you do it using buffer, you edit quickly accessible memory and then save everything in one go.

His example is explaining the costs of operation. For loading memory to data you have 100 operation of 0.5 s cost and then you have one of altering the disc (IO operation) what is not described in the answer and is probably not obvious, nearly all disc provide the bulk transfer alteration operation. So 1 IO operation means 1 save to a disc, not necessarily 1 bit save (it can be much more data).

cerkiewny
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When writing 1 byte at a time, each write requires:

  1. disk seek time (which can vary) to place the 'head' over the correct track on the disk,
  2. disk rotational latency time while waiting for the correct sector of the disk to be under the 'head'.
  3. disk read time while the sector is read, (the rotational latency and sector read time may have to be performed more than once if the CRC does not match that saved on the disk
  4. insert the new byte into the correct location in the sector
  5. rotational latency waiting for the proper sector to again be under the 'head'
  6. sector write time (including the new CRC).

Repeating all the above for each byte (esp. since a disk is orders of magnitude slower than memory) takes a LOT of time.

It takes no longer to write a whole sector of data than to update a single byte.
That is why writing a buffer full of data is so very much faster than writing a series of individual bytes.

There are also other overheads like updating the inodes that:

  • track the directories
  • track the individual file

Each of those directory and file inodes are updated each time the file is updated.

Those inodes are (simply) other sectors on the disk. Overall, lots of disk activity occurs each time a file is modified.

So modifying the file only once rather than numerous times is a major time saving. Buffering is the technique used to minimize the number of disk activities.

Fabricio Araujo
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user3629249
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Among other things, data is written to disk in whole "blocks" only. A block is usually 512 bytes. Even if you only change a single byte inside the block, the OS and the disk will have to write all 512 bytes. If you change all 512 bytes in the block before writing, the actual write will be no slower than when changing only one byte.

The automatic caching inside the OS and/or the disk does in fact avoid this issue to a great extent. However, every "real" write operation requires a call from your program to the OS and probably all the way through to the disk driver. This takes some time. In comparison, writing into a char/byte/... array in your own process' memory in RAM does virtually cost nothing.

JimmyB
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