php - passing argument to function

Question

I have the simple code:

class A {
    public $var = 1;
    function inc() {$this->var++;}
}

function foo($a) {
    for($i = 0; $i < 10; $i++) {
        $a->inc();
    }
}

$a = new A();
foo($a);

$v = $a->var;
echo "var value is $v \n";

I was expecting to get printed the value of 1 but I get 11. shouldn't PHP pass argument to functions with copy-constructor?

Answer 1

PHP does have a copy constructor but it copies shallowly so the object passed to foo is a new copy but it has a reference to $var that points to the same value as $a->var. If you would have done foo(clone($a)) then you would have gotten the answer you're expecting.

Answer 2

You already have $var=1 and you are incrementing every 10 times. That means that your for loop runs 10 times and it gets incremented as you have $this->var++ in class A inc method . So To get your result your $var should be equal to -9 or else you should set the for loop to run single time and set the value of $var as 0 by default.