I'm writing a script to audit updates to files in a set of production unix directories (until we can implement a proper change control process).

I understand that mode from stat() contains file type and permission. As per the user manual, I extract the permissions using %o and a mask (see example below).

Question #1:
How can I extract the file type from mode ?
The mode (5 digit decimal) converts to a 6 digit octal.
Using the mask, returns the permissions (4 digits octal) (see example below).

Question #2:
Is there an easier way to audit changes to files ?
Are there existing functions or scripts to do this ?
(i.e. am I reinventing the wheel ?)

My approach is as follows:
I'll run a nightly cron job which will identify file adds and removes by comparision to previous nights results. I'll identify modifies and changes via mtime and ctime. I'll store all add/remove/update/changes in an audit file which will contain all the info in stat().

Thanks in advance for any help.

Joe

Example:
=======
bshell6.0 > cat test.pl
#!/opt/perl64/bin/perl
my $file = $ARGV[0];
my $mode = (stat($file))[2];
print  "File: $file\n";
print  "mode         = $mode\n";
printf "mode (octal) = %o\n", $mode;
printf "Permissions  = %04o\n", $mode & 07777;
print  "=============================\n";

bshell6.0 > ls -ldn testfile testdir
drwxr-x---   2 643 125 96 May 26 14:24 testdir
-rwxr-x---   1 643 125  0 May 26 14:24 testfile

bshell6.0 > test.pl testfile
File: testfile
mode         = 33256
mode (octal) = 100750
Permissions  = 0750
=============================
bshell6.0 > test.pl testdir
File: testdir
mode         = 16872
mode (octal) = 40750
Permissions  = 0750
=============================

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In reply to Extract file type from stat() mode by joe_s

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