Object property does not seem to work as variable function?

Question

Iam working on sample codes in php, I read the variable functions (which is kind of similar to function pointers in C) and started to write below code. Take a look at this

 class fruit{
     public $run;
 }

 function ano1(){
    // some statements
 }

 $x = 'ano1'; // variable function
 $x(); // works

 $f = new fruit;
 $f->run = 'ano1'; // lets store function to $run property
 $f->run(); // call to undefined method why?

Reason I think because it needs $this reference to be passed implicitly to ano1() which is not a member function? So using $this is not valid inside ano1() they simply did not implement it yet? But Iam a c++ background which is quite possible in c++:

class fruit{ 
   public : void (*run) ();
};

void ano1(){ cout << "im running" << endl; }

int main(){
  fruit f;
  f.run = ano1;
  f.run(); // works as long as you dont use this
}

Or is there a workaround in php as well? (Without closures and anonymous functions). Thank you

Workaround requires the magic `__set()` and `__get()` methods : http://stackoverflow.com/questions/4713680/php-get-and-set-magic-methods ... but only with class members - PHP doesn't really do method overloading as such, see : http://stackoverflow.com/questions/4697705/php-function-overloading and http://www.php.net/manual/en/language.oop5.overloading.php — CD001, Sep 07 '15 at 14:51

deceze · Accepted Answer · 2015-09-07T15:12:19.917

It works entirely differently than function pointers in C.

The particular reason for the error message you're seeing is that PHP cannot tell the difference between $fruit->run() and $fruit->run(); that is, the difference between calling the method run of $fruit, and calling whatever function name $fruit->run evaluates to. And it errs on the wrong side of what you expect. You can disambiguate with call_user_func($fruit->run).

PHP doesn't have anything like "function pointers". Functions are second class citizens in PHP. $x merely holds a string, and PHP's way of supporting passing functions around is to pass them by name. When attempting to "call a string", PHP will instead check if there's a function of that name, and call it instead.

So what you're doing is you're merely assigning a string to an object property, but PHP's syntax is too ambiguous to allow you to call it from there. It is not possible at all _{(with any sort of sane code)} to graft a function onto an object after the fact. All object methods must be defined in the class of the object instance.

score 1 · Answer 2 · answered Sep 07 '15 at 15:03

It's because doing $f->run() says 'call the run method of the $f object'.

Instead, you could do:

$str = $f->run;
$str();

which invokes the string as a function call.
Or

 $f = new fruit;
 $f->run = 'ano1';
 call_user_func($f->run)

which then simply invokes the property run as a method

A way that you could... sort of automate that is by using magic method __call like so:

class Fruit {

    public $run;

    public function __call($methodName,$argsArray){

        if (property_exists($this,$methodName)
                &&is_callable($this->$methodName)){
            return call_user_func_array($this->$methodName,$argsArray);
        } else {
            throw new \Exception("Method {$methodName} does not exist");
        }

    }
}

Then, you would do:

$fruit = new Fruit();
$fruit->run = 'ano1';
$fruit->run();

Then run is not defined in the class Fruit, so it passes the method name (run) to __call with any arguments you passed to run. Then if Fruit has a property named $methodName (in this case 'run'), and the property can be called as a function, then use call_user_func_array to invoke it as a function.

score -1 · Answer 3 · answered Sep 07 '15 at 15:08

-1

Use the below code

$f->{$f->run}();

Since your was treating a run() method on a object of class fruit

answered Sep 07 '15 at 15:08

user3535945

243
1
10

`Call to undefined method fruit::ano1()` – deceze Sep 07 '15 at 15:09
Sorry i assumed that it was inside the class you can use call_user_func($f->run) – user3535945 Sep 07 '15 at 15:12

Object property does not seem to work as variable function?

3 Answers3