in reply to Can caller() or the call stack be tricked?

It looks like a job for goto

After the goto, not even caller() will be able to tell that this routine was called first.

So if you goto &{'debug'}; after you shitfed $self out, you'll "call" your debug subroutine seemlessly, as if you were there in the first place. Be careful about unshifting or shifting the $self argument into @_ when you do that.

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Re^2: Can caller() or the call stack be tricked?
by Ralesk (Pilgrim) on Jul 08, 2013 at 13:09 UTC

    Hah!, I knew I’d seen something like this before — thanks!

    Also, that manual is the most hilarious thing ever. “Create spaghetti code”.

    One of the reasons I love Perl is that it has quips like this throughout its documentation.

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