I would guess that the comma is being used as an operator. Deparse seems to agree: 
    perl -MO=Deparse -e '$x=[1, , ,2]' 
    $x = [1, 2];

[download]
The Comma Operator, is documented in perlop.

Update: My guess was wrong. The comma isn't being used in a scalar context. As an example of that see: 

    perl -MO=Deparse -e '$x=(1, , ,2)' 
    $x = ('???', 2);

[download]
merlyn seems to have the correct reason.

--
John.

In reply to Re: Bug with lists? by jmcnamara
in thread Bug with lists? by demerphq

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