PHP5.3: "Call to undefined method" error when calling invoke from class variable

Question

I have been doing some tests (to replace old code) with the __invoke magic method and I'm not sure this is a bug or not:

Lets suppose we have a class:

class Calc {
    function __invoke($a,$b){
        return $a*$b;
    }
}

The following is possible and works without any problem:

$c = new Calc;
$k = $c;
echo $k(4,5); //outputs 20

However if I want to have another class to store an instance of that object, this doesn't work:

class Test {
    public $k;
    function __construct() {
        $c = new Calc;
        $this->k = $c; //Just to show a similar situation than before
        // $this-k = new Calc; produces the same error.
    }
}

The error occurs when we try to call it like:

$t = new Test;
echo $t->k(4,5); //Error: Call to undefined method Test::k()

I know that a "solution" could be to have a function inside the class Test (named k) to "forward" the call using call_user_func_array but that is not elegant.

I need to keep that instance inside a common class (for design purposes) and be able to call it as function from other classes... any suggestion?

Update:

I found something interesting (at least for my purposes):

If we assign the "class variable" into a local variable it works:

$t = new Test;
$m = $t->k;
echo $m(4,5);

Please provide a complete code example which shows your problem. The code sample you say works does not work and the code sample you say gives a problem works. — Sjoerd
– Sjoerd, Commented Jun 24, 2010 at 7:54
Yes, sorry, my bad... I update it. Sjoerd: Did that example worked for you? strange :S — lepe
– lepe, Commented Jun 24, 2010 at 7:56

Sjoerd · Accepted Answer · 2010-06-24 07:59:16Z

7

PHP thinks you want to call a method k on instance $t when you do:

$t->k(4, 5)

which is perfectly reasonable. You can use an intermediate variable to call the object:

$b = $t->k;
$b(4, 5);

See also bug #50029, which describes your issue.

answered Jun 24, 2010 at 7:59

Sjoerd

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1 Comment

lepe Over a year ago

(+1). Now it is getting clear for me. Thank you for the link.

Community · Accepted Answer · 2023-11-17 19:32:19Z

6

When you do $test->k(), PHP thinks you are calling a method on the $test instance. Since there is no method named k(), PHP throws an exception. What you are trying to do is make PHP return the public property k and invoke that, but to do so you have to assign k to a variable first. It's a matter of dereferencing.

You could add the magic __call method to your Test class to check if there is a property with the called method name and invoke that instead though:

public function __call($method, $args) {
    if(property_exists($this, $method)) {
        $prop = $this->$method;
        return $prop();
    }
}

I leave adding the arguments to the invocation to you. You might also want to check if the property is_callable.

But anyway, then you can do

$test->k();

edited Nov 17, 2023 at 19:32

CommunityBot

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answered Jun 24, 2010 at 7:57

Gordon

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1 Comment

lepe Over a year ago

We posted at the same time... Yes maybe that could help... If I don't find any other better alternative I will mark your answer as good.

StasM · Accepted Answer · 2010-06-24 08:08:34Z

2

You can not use method syntax (like $foo->bar() ) to call closures or objects with __invoke, since the engine always thinks this is a method call. You could simulate it through __call:

function __call($name, $params) {
  if(is_callable($this->$name)) {
    call_user_func_array($this->$name, $params);
  }
}

but it would not work as-is.

answered Jun 24, 2010 at 8:08

StasM

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Comments

David Ginanni · Accepted Answer · 2019-10-19 08:55:40Z

0

If you call $test->k() PHP will search for a method called "k" on the $test instance and obviously it will throws an Exception.

To resolve this problem you can create a getter of the property "k"

class Test {
    public $k;

    function __construct() {
        $c = new Calc;
        $this->k = $c; //Just to show a similar situation than before
        // $this-k = new Calc; produces the same error.
    }

    public function getK() {
        return $this->k;
    }
}

So now you can use the functor in this way:

$t = new Test();
echo $t->getK()(4,5);

edited Oct 19, 2019 at 8:55

answered Oct 16, 2019 at 22:14

David Ginanni

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