No, 0||$x is just $x, not "Booleanized" $x. perl -le"print 0||5" prints 5 not 1 and perl -le"print 1&&@ARGV" x y z prints xyz.

- tye        

In reply to Re^5: One out of three ain't bad (! !!) by tye
in thread One out of three ain't bad by saintmike

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