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?

Answer 1

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.

Answer 2

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.)

Answer 3

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.