Wrong. That's equivalent to 
    $foo = $bar unless defined $foo;

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What the OP wants is: 
    $foo = $bar // $foo;

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which is equivalent to 
    $foo = $bar if defined $bar;

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assuming the absence of ties and overloading.

In reply to Re^2: Assignment to a value only if it is defined by JavaFan
in thread Assignment to a value only if it is defined by voj

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