2

There is a concept in bash called process substitution. You can run a command and use the output as a file.

e.g.

$ cat -n <(seq 3|tac)
     1  3
     2  2
     3  1

I am curious why the following fails;

$ du -sk <(xzcat /var/log/mpd/scribble.log.xz )
0   /dev/fd/63

Similarly, we have this

$ file <(seq 1 2)
/dev/fd/63: broken symbolic link to pipe:[32560687]

The file is not empty.

$ xzcat /var/log/mpd/scribble.log.xz  | wc -c 
16877047

>/dev/null  pv  <(xzcat /var/log/mpd/scribble.log.xz ) 
16.1MiB 0:00:00 [ 232MiB/s] [  <=>                   ]

As a bonus question, do you know a technique in bash to masquarade fifos as regular files for a short duration?

Ярослав Рахматуллин
  • 11,633

1 Answers1

2

Regular files consist of three main constituents:

  1. Name (hardlink)
  2. Inode (system info)
  3. Body (data)

Pipes (no matter named or anonymous, persistent or temporary) have no body. Therefore they always have zero size unlike regular files. They only have in-memory buffer. That's why anonymous named pipe /dev/fd/63 has zero length. You could use any other command within <(...) operator with the same result, say

du -sk <(cat somefile)

You can create named pipe with mkfifo command to see that its size is always equal to zero even if you fully fill its buffer. Actually, filesize of named (and any other) pipes has no real sense and is set to zero for uniformity of listing among other files.

Oleg Bolden
  • 1,735