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Re: $^E bug and perl 5.8.0

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Dan Sugalski
November 19, 2003 07:54
Re: $^E bug and perl 5.8.0
Message ID:
On Wed, 19 Nov 2003, Chris Nandor wrote:

> $ perl -le '$^E = -1728; print $^E+0 for 0,1; print $!+0 for 0,1'
> -1728
> 22
> 22
> 22

Given that on Unix systems $^E and $! are identical, and that $! is
directly tied to errno, this certainly isn't surprising. Anything that
afects errno (like, say, IO) may then affect $! and $^E identically.(And
$!/$^E non-indentically on systems where they're separate)

Basically you're using a variable that can be affected by external things
in a way that pretty much guarantees that external things will be
happening. That it changes isn't much of a surprise. ($!/$^E may get
modified by some signal handlers too, depending on what you do, so there's
not even any guarantee that "$^E = 42; $^E++;" will end up with $^E set to


--------------------------------------"it's like this"-------------------
Dan Sugalski                          even samurai                         have teddy bears and even
                                      teddy bears get drunk

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