Unix pipe into ls

Question

I thought I understood *nix pipes until now... I have an executable called studio which symlinks to my install of Android Studio and I had assumed I could get the linked-to location with

which studio | ls -l

But that doesn't work. What it gives me is equivalent to having just run ls -l in the current directory.

If I run which studio, I get /home/me/bin/studio. And if I run ls -l /home/me/bin/studio I get the expected output showing me the symlink location.

So why doesn't the piped version work? What haven't I grokked about pipes?

score 97 · Accepted Answer · edited May 23 '17 at 11:54

97

To do that you need xargs:

which studio | xargs ls -l

From man xargs:

xargs - build and execute command lines from standard input

To fully understand how pipes work, you can read What is a simple explanation for how pipes work in BASH?:

A Unix pipe connects the STDOUT (standard output) file descriptor of the first process to the STDIN (standard input) of the second. What happens then is that when the first process writes to its STDOUT, that output can be immediately read (from STDIN) by the second process.

edited May 23 '17 at 11:54

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answered Aug 13 '13 at 13:51

fedorqui

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This does not seem to work when I try it with ls without arguments. So `which studio | xargs ls` just outputs the result of which for me. Why does it require arguments to work? – bdbdbd Aug 06 '23 at 12:21
@bdbdbd probably because `ls $(which studio)` is exactly the same. `ls`-ing the output of `which` is not going to give you anything else than a route, which is what `ls` gives as well. – fedorqui Aug 11 '23 at 20:23

score 33 · Answer 2 · answered Aug 13 '13 at 13:55

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ls does not read its arguments from standard input, but from the command line. To get the directory at the command line, you have to use command substitution:

ls -l "$( which studio )"

(The double quotes are needed if the path might contain whitespace.)

answered Aug 13 '13 at 13:55

choroba

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The answer involving `xargs` is cleaner, but this honors my shell's color coding, which is a nice boon! – Dannid Oct 15 '14 at 16:48
@Dannid when I tried xargs I got `Too many errors from stdin` , but this answer worked to me out the box . – Salem Oct 05 '18 at 20:19
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How could I know if a command reads its arguments from stdin or command line please? – Tengerye Nov 07 '21 at 02:23
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@Tengerye: Read its documentation. – choroba Nov 07 '21 at 18:16
@Tengerye `man command`, in the head of the description section. If nothing says "read standard input" then it probably can't. – Glastis Jun 06 '23 at 08:09

score 4 · Answer 3 · answered Oct 31 '16 at 18:17

Since ls -l does not take any input, it does not do anything regarding the output of which studio. The important thing here is understanding the difference between standard input and arguments. The standard input is a special file that is read using the scanf procedure (by a program in C for example), and the arguments to a program are passed to the main procedure as the argv and argc parameters. argv is an array of null terminated arrays of char, and argc is the length of that array.

Unix pipe into ls

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