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Date: Mon, 27 Sep 2004 11:52:24 -0600
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Subject: return NULL by reference
From: jeff@opendbms.com ("Jeffrey Moss")

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I can't seem to reproduce this bug, I was wondering if anyone has any =
input they could give me....

I wrote a function to return a parameter by reference that worked with =
PHP5.0 RC1, looks like this:

 function & getParameter ($name)=20
    {=20
        if (isset($this->params[$name]))=20
        {=20
            return $this->params[$name];=20
        }=20
        return NULL;=20
    }=20

Now I recently upgraded to PHP5.0.2 and I get errors. I have tried to =
make a test case script to demonstrate it but it works fine in command =
line. My cgi script is huge, so I can't tell you how to reproduce it, =
but it's giving me this error:  "Only variable references should be =
returned by reference " on the line with "return NULL;".

Could this be related to another bug? The code works fine if I replace =
"return NULL" with anything else, even "return $nonexistant->variable;". =
Is there a way I can turn this check off?

-Jeff Moss
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