in reply to Re: Use of uninitialized value in open second time but not first.
in thread Use of uninitialized value in open second time but not first.
So it actually spits a warning the first time.
Where does that warning go? It's not on my screen when I run it. As I see it, $output is uninitialized both times, but the warning is not printed the first time.
I'm confused also because what I pass to open is a reference to $output which is always a real value regardless of whether $output is initialized or not.
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Re^3: Use of uninitialized value in open second time but not first.
by Bro. Doug (Monk) on Apr 01, 2007 at 16:29 UTC |