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I'm writing a script that needs canonical path of certain commands. Since there could be symbolic links pointing to the actual commands, I use readlink -f to get the canonical path. But I'm not getting what I actually want with readlink -f, I'll explain with the following example:

Let's say my current directory is: /home/user/Documents If I try to get a path of sleep with readlink -f I get this:

/home/user/Documents/sleep

What I actually want is /bin/sleep

  • Does sleep (either a file or a symlink) actually exist in that directory? – JigglyNaga May 20 '16 at 12:03
  • No, they are mostly commands (not in the script directory) – mythic May 20 '16 at 12:58
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readlink -f "$(type -P sleep)"

or if you're performance-conscious:

cpath="$(type -P sleep)"; [ ! -L "$cpath" ] || cpath="$(readlink -f "$cpath")"

Using readlink -e (existing) instead of readlink -f can save you from this kind of accident where you operate on a nonexisting file.

The second example assumes the path returned by type -P is canonical, which means it assumes your path doesn't have non-canonical components.

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    [ ! -L "$cdpath" ] is not a guarantee that the path is canonical or even absolute. The path could still be relative, and another of the components but the last may be a symlink. – Stéphane Chazelas May 20 '16 at 12:06
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    Beware which has a number of issues. The standard command would be command -v (though would return the command name for those commands that are builtin). – Stéphane Chazelas May 20 '16 at 12:07

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