What does “< <(command args)” mean in the shell?

痴心易碎 提交于 2021-02-04 09:43:55

问题


When looping recursively through folders with files containing spaces the shell script I use is of this form, copied from the internet:

    while IFS= read -r -d $'\0' file; do
      dosomethingwith "$file"        # do something with each file
    done < <(find /bar -name *foo* -print0)

I think I understand the IFS bit, but I don't understand what the '< <(...)' characters mean. Obviously there's some sort of piping going on here.

It's very hard to Google "< <", you see.


回答1:


<() is called process substitution in the manual, and is similar to a pipe but passes an argument of the form /dev/fd/63 instead of using stdin.

< reads the input from a file named on command line.

Together, these two operators function exactly like a pipe, so it could be rewritten as

find /bar -name *foo* -print0 | while read line; do
  ...
done



回答2:


<( command ) is process substitution. Basically, it creates a special type of file called a "named pipe," then redirects the output of the command to be the named pipe. So for example, suppose you want to page through a list of files in an extra-big directory. You could do this:

ls /usr/bin | more

Or this:

more <( ls /usr/bin )

But NOT this:

more $( ls /usr/bin )

The reason for this becomes clear when you investigate further:

~$ echo $( ls /tmp )
gedit.maxtothemax.436748151 keyring-e0fuHW mintUpdate orbit-gdm orbit-maxtothemax plugtmp pulse-DE9F3Ei96ibD pulse-PKdhtXMmr18n ssh-wKHyBU1713 virtual-maxtothemax.yeF3Jo
~$ echo <( ls /tmp )
/dev/fd/63
~$ cat <( ls /tmp )
gedit.maxtothemax.436748151
keyring-e0fuHW
mintUpdate
orbit-gdm
orbit-maxtothemax
plugtmp
pulse-DE9F3Ei96ibD
pulse-PKdhtXMmr18n
ssh-wKHyBU1713
virtual-maxtothemax.yeF3Jo

/dev/fd/whatever acts like a text file with the output of the command between the parenthesis.




回答3:


< redirects to stdin.

<() seems to be some sort of a reverse pipe, as mentioned on the page:

find /bar -name *foo* -print0 | \
while IFS= read -r -d $'\0' file; do
  dosomethingwith "$file"        # do something with each file
done

will not work, because the while loop will be executed in a subshell, and you'll lose changes made in the loop



来源:https://stackoverflow.com/questions/2443085/what-does-command-args-mean-in-the-shell

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