linux - why is the output of `du` often so different from `du -b`

Question

why is the output of du often so different from du -b? -b is shorthand for --apparent-size --block-size=1. only using --apparent-size gives me the same result most of the time, but --block-size=1 seems to do the trick. i wonder if the output is then correct even, and which numbers are the ones i want? (i.e. actual filesize, if copied to another storage device)

question from:https://stackoverflow.com/questions/5694741/why-is-the-output-of-du-often-so-different-from-du-b

Answer 1

Apparent size is the number of bytes your applications think are in the file. It's the amount of data that would be transferred over the network (not counting protocol headers) if you decided to send the file over FTP or HTTP. It's also the result of cat theFile | wc -c, and the amount of address space that the file would take up if you loaded the whole thing using mmap.

Disk usage is the amount of space that can't be used for something else because your file is occupying that space.

In most cases, the apparent size is smaller than the disk usage because the disk usage counts the full size of the last (partial) block of the file, and apparent size only counts the data that's in that last block. However, apparent size is larger when you have a sparse file (sparse files are created when you seek somewhere past the end of the file, and then write something there -- the OS doesn't bother to create lots of blocks filled with zeros -- it only creates a block for the part of the file you decided to write to).