unexpected '[', expecting ')' - array in constructor

Question

I have this code:

class foo
{
    public function __construct
    (
        $bar[var1]  = 0,
        $bar[var2]  = 0,
        $bar[var3]  = 0
    )
    {
        /* do something */
    }
}

and I get this error:

Parse error: syntax error, unexpected '[', expecting ')' in $this on line 5

and I have no idea about whats wrong.

EDIT:

I want to initialize the class like this: $instance = new foo($bar) ($bar is an array)

Because the class would expect many arguments, and I can easily return the array from the constructor. Let me know if this is not a good idea.

this being the EXACT method of giving space in actual code? — hjpotter92
– hjpotter92, Commented Apr 26, 2012 at 8:47
What are you trying to do? For instance, how are you planning to call this function? — Mr Lister
– Mr Lister, Commented Apr 26, 2012 at 8:48

ThiefMaster · Accepted Answer · 2012-04-26 10:38:43Z

8

Why are you trying to create an array in the function argument definition? That makes no sense and you cannot do this.

If you want $bar to be an argument but ensure some values exist (i.e. default values), do it like this:

public function __construct($bar)
{
    if(!isset($bar['var1'])) $bar['var1'] = 0;
    if(!isset($bar['var2'])) $bar['var2'] = 0;
    if(!isset($bar['var3'])) $bar['var3'] = 0;

    /* do something */
}

or in a more compact way:

public function __construct($bar)
{
    $bar = array_merge(array('var1' => 0, 'var2' => 0, 'var3' => 0), $bar);
    /* do something */
}

edited Apr 26, 2012 at 10:38

answered Apr 26, 2012 at 8:46

ThiefMaster

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1 Comment

DaveRandom Over a year ago

Shorter way of writing it without the ifs: $bar = array_merge(array('var1'=>0, 'var2'=>0, 'var3'=>0), $bar);. But +1.

Zheng Kai · Accepted Answer · 2012-04-26 08:47:07Z

1

class foo
{
    public function __construct($bar = array())
    {
        $bar += array(
            "var1" => 0,
            "var2" => 0,
            "var3" => 0
        );

        /* do something */
    }
}

answered Apr 26, 2012 at 8:47

Zheng Kai

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Comments

Gustav Westling · Accepted Answer · 2012-04-26 08:47:56Z

1

class foo
{
    public function __construct()
    {
        $bar = func_get_args();
    }
}

answered Apr 26, 2012 at 8:47

Gustav Westling

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Comments

Christophe · Accepted Answer · 2012-04-26 08:48:20Z

1

you cannot do that you should pass the array as a whole. also the keys have to be quoted

public function __construct($bar)
    {
        $bar['var1']  = 0;
        $bar['var2']  = 0;
        $bar['var3']  = 0;

        /* do something */
    }

answered Apr 26, 2012 at 8:48

Christophe

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Comments

PeeHaa · Accepted Answer · 2012-04-26 08:49:11Z

1

class foo
{
    public function __construct($bar = array('var'=>0, 'var2'=>0, 'var3'=>0))
    {
        var_dump($bar);
    }
}

answered Apr 26, 2012 at 8:49

PeeHaa

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