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state and file perms confusion

by neilwatson (Priest)
on Aug 04, 2012 at 01:06 UTC ( #985371=perlquestion: print w/replies, xml ) Need Help??
neilwatson has asked for the wisdom of the Perl Monks concerning the following question:

Greetings Sisters and Brothers, Using stat, I'm not getting the results I expect. Could you please explain why.
($dev,$ino,$mode,$nlink,$uid,$gid,$rdev,$size, $atime,$mtime,$ctime,$blksize,$blocks) = stat($file); ... Later $mode = $mode & 07777; print "mode=".$mode."\n";
In this example the target file is /etc/passwd. Here are my results.
$ ls -l /etc/passwd -rw-r--r--. 1 root root 1289 Jul 22 16:27 /etc/passwd $ ./ /etc/passwd 644 root root mode=420
Why is the mode 420 rather than 644?

Neil Watson

Replies are listed 'Best First'.
Re: state and file perms confusion
by Kenosis (Priest) on Aug 04, 2012 at 01:20 UTC

    From perldoc:

    Because the mode contains both the file type and its permissions, you +should mask off the file type portion and (s)printf using a "%o" if y +ou want to see the real permissions. $mode = (stat($filename))[2]; printf "Permissions are %04o\n", $mode & 07777;

    You've masked off the file type with $mode = $mode & 07777;, so you'll get 0644 (octal) from the script above when your $mode equals 420 (decimal), since the printf above makes the conversion from decimal to octal.

      I'm missing something fundamental.
      print "mode=".$mode."\n"; printf "printf are %04o\n", $mode; printf "Permissions are %04o\n", $mode & 07777; ..produces mode=33188 printf are 100644 Permissions are 0644
      What am I seeing? How can store the number in a scalar rather than printing it?

      Neil Watson

        What am I seeing?

        33188 decimal = 100644 octal (first printf) 07777 octal = 0000111111111111 binary (mask)-| +- & (bitwise AND) 33188 decimal = 1000000110100100 binary (mode)-| 0644 octal = 0000000110100100 binary (printf after mask) ^^^^ | + - file type masked out

        How can store the number in a scalar rather than printing it?

        use Modern::Perl; my $mode = 33188; my $modeResult = sprintf '%04o', $mode & 07777; say $modeResult; # prints 0644

        There aren't a lot of use cases for storing an octal "number" in a scalar. There are some, but they're all to do with formatting output, not actual use as a number.

        If your goal is to use this for chmod, you have to realise that it won't work. If you do this:

        $mode = sprintf "%o", 420; # $mode = 644 chmod $mode, $file;
        you'll actually be doing the same thing as this:
        chmod 01204, $file; # 01204 octal == 644 decimal
        Definitely not what you wanted. Leave the number as-is if you plan on using it in chmod. Note that the following lines are equivalent:
        # 1. octal number chmod 0644, $file; # 2. decimal number chmod 420, $file; # 3. number in a scalar $mode = 0644; chmod $mode, $file; # 4. number in a scalar (from decimal) $mode = 420; chmod $mode, $file; # 5. number with bitmasks $mode = 0666 & ~022; # like umask chmod $mode, $file; # 6. decimal with bitmasks $mode = 438 & ~022; # 438 == 0666 chmod $mode, $file;
        So, to answer your question, the number is already in the scalar. You don't actually need to do anything about it.

        And, if you use $mode = sprintf "%o", $mode;, remember what you just did: you took a number, you printf'd it into a string, and put that back in $mode. So $mode no longer carries a number but a string. Perl will try to force it back to a number if that's what you try to do, but it won't be the number you started with.

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