Can't call static method from class as variable name?

Question

I'm using php 5.2.6. I have a strategy pattern, and the strategies have a static method. In the class that actually implements one of the strategies, it gets the name of the strategy class to instantiate. However, I wanted to call one of the static methods before instantiation, like this:

$strNameOfStrategyClass::staticMethod();

but it gives T_PAAMAYIM_NEKUDOTAYIM.

$> cat test.php

<?

interface strategyInterface {
        public function execute();
        public function getLog();
        public static function getFormatString();
}


class strategyA implements strategyInterface {
        public function execute() {}
        public function getLog() {}
        public static function getFormatString() {}
}

class strategyB implements strategyInterface {
        public function execute() {}
        public function getLog() {}
        public static function getFormatString() {}
}

class implementation {
        public function __construct( strategyInterface $strategy ) {
                $strFormat = $strategy::getFormatString();
        }
}

$objImplementation = & new implementation("strategyB") ;

$> php test.php

Parse error: syntax error, unexpected T_PAAMAYIM_NEKUDOTAYIM in /var/www/test.php on line 24

$> php -v

PHP 5.2.6-1+lenny9 with Suhosin-Patch 0.9.6.2 (cli) (built: Aug  4 2010 03:25:57)

Would this work in 5.3?

ircmaxell · Accepted Answer · 2010-09-24T14:34:31.433

Yes. That syntax was introduced in 5.3

To workaround for <= 5.2, you can use call_user_func:

call_user_func(array($className, $funcName), $arg1, $arg2, $arg3);

or call_user_func_array:

call_user_func_array(array($className, $funcName), array($arg1, $arg2, $arg3));

But on another note, what you're trying to do doesn't really make sense...

Why have it as a static function? Your constructor in implementation is expecting an object anyway (that's what strategyInterface $strategy is looking for). Passing a string won't work, since strings don't implement interfaces. So what I would do, is make the interface non-static, and then do something like:

$strategy = new StrategyB();
$implementation = new Implementation($strategy);

Then, in the constructor:

$strFormat = $strategy->getFormatString();

Or, if you really still want that method to be static you could do:

$strFormat = call_user_func(array(get_class($strategy), 'getFormatString'));

Oh, and = & new synax is deprecated (and doesn't do what you think it does anyway).

Having a brain fart today :P I don't want to pass an instantiated object to the implementing class, because I want to instantiate it on down the line, after the implementing class has created some data that the strategy needs. I'll think I'll need the `call_user_func_array()` anyway -- thanks for the great code! — user151841, Sep 24 '10 at 15:02
Note that this only sort of works in 5.3. You can do "$foo::staticMethod()" but you cannot do "$this->foo = new Whatever(); $this->foo::staticMethod()". — , Dec 21 '12 at 23:37

score 0 · Answer 2 · answered Sep 24 '10 at 14:27

0

Type hinting is going to give you some problems:

Argument 1 passed to implementation::__construct() must implement interface strategyInterface, string given

answered Sep 24 '10 at 14:27

Mark Baker

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Hm, brain fart this morning. Why didn't my php give me that error? – user151841 Sep 24 '10 at 14:55
@user151841 - Because unexpected T_PAAMAYIM_NEKUDOTAYIM is detected at compile time, so you were getting that error... I fixed that with call_user_func() before even trying to run the code, and got the second error, which is a runtime error. – Mark Baker Sep 24 '10 at 15:17

Can't call static method from class as variable name?

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