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Re: Determining a module's user

by Fletch (Chancellor)
on Oct 10, 2007 at 16:37 UTC ( #644024=note: print w/replies, xml ) Need Help??

in reply to Determining a module's user

caller looks to work sanely inside the Filter::Simple called block (at least with a quick test module).

$ cat package FSTest; use Filter::Simple; FILTER { print STDERR "in FILTER, caller package: ", (caller( 1 ))[0], "\n"; } 1; $ perl -MFSTest -e 0 in FILTER, caller package: main

Update: Oop. Quick test module FTL then apparently. You may can get at it by calling caller in a custom import sub of your own, of you might have to fall back to using Filter::Util::Call instead. We're past my simple experience with Filter::Simple at any rate.

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Re^2: Determining a module's user
by clueless newbie (Chaplain) on Oct 10, 2007 at 17:18 UTC
    For me (Windows/Activestate 5.8) it doesn't. If "main" calls "package" and "package" uses the filtering module, (caller( 1 ))[0] returns not "package" but "main".

    I've found a poor work around that involves "use myFilter __PACKAGE__;".

    Thanks, Fletch!
    --- it seems to be (caller(1))[1]!

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