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Re: ||= (poorly documented?)

by kcott (Canon)
on Jul 09, 2012 at 06:02 UTC ( #980653=note: print w/ replies, xml ) Need Help??


in reply to ||= (poorly documented?)

In perlop - Assignment Operators, the example shown for += applies to all the op= operators. So,

$x ||= 2;

is equivalent to

$x = $x || 2;

-- Ken

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