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.

In reply to Re: Can caller() or the call stack be tricked? by Eily
in thread Can caller() or the call stack be tricked? by Ralesk

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