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Increment on "__toString"

I am not sure what the title should be, but the code should explain it better:

class Group {
    private $number = 20;

    public function __toString() {
        return "$this->number";
    }
}

$number = new Group();
echo $number, PHP_EOL;
echo ++ $number, PHP_EOL;

echo PHP_EOL;

$number = "20";
echo $number, PHP_EOL;
echo ++ $number, PHP_EOL;

echo PHP_EOL;

$number = 20;
echo $number, PHP_EOL;
echo ++ $number, PHP_EOL;

Output:

20
20              <--- Expected 21

20
21

20
21

Any idea why I got 20 instead of 21? Even then the code below works:

$i = null ;
echo ++$i ; // output 1

I know Group is an object that implements __toString , i expected ++ to work with the string from __toString or at least throw an error

like image 210
Baba Avatar asked May 04 '13 14:05

Baba


4 Answers

The order in which the operations happen is important:

  1. The variable will be fetched as an object, it won't be casted to an integer (or something else).

  2. This ++ operator increments the lval (the long value) of the zval, but does normally nothing else. The object pointer remains the same. The internal (fast_)increment_function will be called with the zval which has a pointer to the object, which checks for the type first. If it's an object, it does nothing. So when your zval is an object, it is as useful as a no-operation. This won't output any warning.

  3. Only then the echo instruction performs a string cast on his arguments: The __toString method is called and returns 20.

  4. 20 will be output.

like image 68
bwoebi Avatar answered Nov 17 '22 17:11

bwoebi


To answer you question with a little bit of code.

$number = new Group();
echo gettype($number);

$number = "20";
echo gettype($number);

$number = 20;
echo gettype($number);

Will result in

object
string
integer

The three cases:

  • You can't do any integer operation on a object, that why your code does not do what you expect. The __toString method will called very late, when the acutal output will computed, after you unsuccessfully tried to do an math operation with it.
  • You can to math with strings, because PHP internally converts them back to numbers
  • Obviously you can do math with integer

Bonus:

This will work:

$number = new Group();
echo 1 + "$number"; // 21

It converts you object into a string, which could be converted into a number for a math operation.

like image 32
TheHippo Avatar answered Nov 17 '22 17:11

TheHippo


I think it might be clearer with just changing the names of the variables like this :

class Group {
    private $number = 20;

    public function __toString() {
        return "$this->number";
    }
}

$group = new Group();
echo $group;//print 20 as per your __toString function

++ $group;

Now it seems obvious : what is supposed to do a '++' operator on a object of type group ??

like image 2
Frédéric Clausset Avatar answered Nov 17 '22 17:11

Frédéric Clausset


Why don't you just:

class Group {
    private $number = 0;
    public function __construct($number = 0){
        $this->number = intval($number);
    }
    public function __toString() {
        return number_format(++$this->number); // pre-increment
    }
}
$g = new Group();
echo $g; // 1
echo $g; // 2

I use something like this to format offsets in tables.

like image 1
CodeAngry Avatar answered Nov 17 '22 16:11

CodeAngry