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== How can I get the permissions of a file without parsing ls -l output? ==

There are several potential ways, most of which are system-specific.  They also depend on precisely ''why'' you want the permissions.

The majority of the cases where you might ask this question -- such as ''I want to find any files with the setuid bit set'' -- can be answered by the information in [:UsingFind#permissions:].  As the page name implies, those answers are based on the `find(1)` command.

For some questions, such as ''I want to make sure this file has 0644 permissions'', you don't actually need to ''check'' what the permissions are.  You can just use `chmod 0644 myfile` and set them directly.

If your needs aren't met by any of those, then we can look at a few alternatives:

 * On GNU/Linux systems, and possibly others, there is a command called `stat(1)`.  On older GNU/Linux systems, this command take no options -- just a filename -- and you will have to parse its output.
 {{{$ stat /
   File: "/"
   Size: 1024         Filetype: Directory
   Mode: (0755/drwxr-xr-x)         Uid: (    0/    root)  Gid: (    0/    root)
 Device:  8,0   Inode: 2         Links: 25   
 Access: Wed Oct 17 14:58:02 2007(00000.00:00:01)
 Modify: Wed Feb 28 15:42:14 2007(00230.22:15:49)
 Change: Wed Feb 28 15:42:14 2007(00230.22:15:49)}}}
 In this case, one could extract the 0755 from the `Mode:` line, using `awk` or similar commands.

 * On newer GNU/Linux systems, the `stat` command takes arguments which allow you to specify which information you want:
 {{{$ stat -c %a /
 755}}}
 That's obviously a lot easier to parse.

 * On systems with perl 5, you can use:
 {{{
 perl -e 'printf "%o\n", 07777 & (stat $ARGV[0])[2]' "$filename"}}}
 This returns the same octal string that the `stat -c %a` example does, but is far more portable.  (And slower.)
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