Increment on "__toString"

Question

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

Does PHP offer a way to overload the increment operators? Since your class doesn't easily convert to a number type, you probably have to tell PHP what you want ++ to do. — dlp, May 04 '13 at 14:34
if you declare $number not private in the Group class and then try echo ++ $number->number, PHP_EOL; works? — LuckyStarr, May 04 '13 at 14:37
Nope .. but that was increment on string and even if otherwise .. i should get an error — Baba, May 04 '13 at 14:37
Prefix ops like `++` work on variables. But `$number` is an expression in case 1, not a variable` — mario, May 04 '13 at 14:38
@mario then it should be treated as a bug ... I should at least get a warning or notice just like when you try to use `int` with `__toString` — Baba, May 04 '13 at 14:47

bwoebi · Accepted Answer · 2013-06-15T10:26:02.933

The order in which the operations happen is important:

The variable will be fetched as an object, it won't be casted to an integer (or something else).
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.
Only then the echo instruction performs a string cast on his arguments: The __toString method is called and returns 20.
20 will be output.

score 9 · Answer 2 · answered May 04 '13 at 14:36

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.

Adding explicitly one to the variable expression actually opens the way to changing the amount added very easy. Pre- or post-increment operations should be avoided. You'll never really know what will happen. [Douglas Crockford doesn't like them.](http://stackoverflow.com/questions/971312/why-avoid-increment-and-decrement-operators-in-javascript) - and in PHP it's way easier to make use of the short version `$i+=1^` instead of `$i++` - just one more character. And try to increment by 2... :) — Sven, Jun 22 '13 at 20:22

score 2 · Answer 3 · answered May 04 '13 at 14:43

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 ??

CodeAngry · Answer 4 · 2013-06-25T02:33:09.927

1

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.

edited Jun 25 '13 at 02:33

answered Jun 24 '13 at 14:25

CodeAngry

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score 0 · Answer 5 · answered Nov 08 '13 at 04:48

This is actually more feasible than you may think -- just need to do a little bit more type casting, as follows:

<?php

class Group {
private $number = 20;

    public function __toString() {
        return (string) $this->number; // replace "" w/string cast
    }
}

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

You don't have to use the string cast of course in the magic __toString() but I personally prefer to read the code that way rather than seeing the quotes -- but I think that's just a stylistic preference.

Casting the newly created object as a string causes the magic __toString method to automatically execute and it returns a numeric string which when cast to an int allows you to display the number, increment it and display it again.

Incidentally, the space between ++ and $number is okay; I closed it up b/c that is what I'm used to in other languages like C.

Increment on "__toString"

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