in reply to qx and $!
If you read perlvar, you'll see about $!:
If used numerically, yields the current value of the C "errno" variable, or in other words, if a system or library call fails, it sets this variable. This means that the value of $! is meaningful only immediately after a failure:
For your purposes, the key phrase is "meaningful only immediately after a failure".
So you're trying to attach meaning to something that isn't useful in this case (ie. the value of errno).
Update: Hey, no fair davorg! You clearly have the perldocs memorized, whereas I had to go look it up; that's why you beat my answer! :)
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Re^2: qx and $!
by blazar (Canon) on Oct 18, 2006 at 16:19 UTC | |
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Re^2: qx and $!
by bart (Canon) on Oct 18, 2006 at 18:38 UTC |