static member cannot be used as a default function parameter in PHP5?

Question

dose php5 forbid this usage, the following code doesn't work

class Foo{
    public static $data = "abcd";
}

function tt($para = Foo::$data)
{
    echo $para;
}

tt ("rcohu");

it reports:

PHP Parse error:  syntax error, unexpected T_VARIABLE, expecting T_STRING in /home/jw/sk/sk.php on line 6

Parse error: syntax error, unexpected T_VARIABLE, expecting T_STRING in /home/jw/sk/sk.php on line 6

if you are happy with my answer, can you mark it as such

Jason
– Jason

2011-01-12 03:34:42 +00:00
Commented Jan 12, 2011 at 3:34 — Jason
– Jason, Commented Jan 12, 2011 at 3:34

Jason · Accepted Answer · 2011-01-10 03:42:50Z

6

function tt($para = Foo::$data)
{
    echo $para;
}

function definitions can only contain simple assignments, not complex ones like Foo::$data.

Just do this:

function tt($para = false)
{
    if(!$para) $para = Foo::$data;
    echo $para;
}

answered Jan 10, 2011 at 3:42

Jason

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2 Comments

BoltClock Over a year ago

They just cannot contain variable expressions. Foo::DATA where DATA is a class constant works.

Jason Over a year ago

True, so if you had define('DATA', 'my-data'); you could use function tt($para=DATA){

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