Assigning an array inside a while loop causes infinite loop

Question

$pie = array('Apple' => 1, 'Cherry' => 3, 'Pumpkin' => 6);
reset($pie);
while (list($k, $v) = each($pie)) {
    echo "$k => $v\n";
    $desserts[]=$pie;
}

If the $desserts[]=$pie line is commented out, this code block executes as expected.

Why when it is uncommented does it go into an infinite loop?

I do not understand why that happens. This is a perfect example of what NOT to do, but I don't understand why the complete array is printed over and over. I thought just the first slice would be repeated.

Community · Accepted Answer · 2023-11-17 20:38:18Z

6

In the manual, it warns you not to do what you're doing.

Caution: Because assigning an array to another variable resets the original arrays pointer, our example above would cause an endless loop had we assigned $fruit to another variable inside the loop.

See documenation for each().

You can achieve the same results using array_fill if you just wanted to make many copies of $pie.

 $desserts = array_fill(0,count($pie),$pie);

edited Nov 17, 2023 at 20:38

CommunityBot

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answered Nov 22, 2013 at 15:48

Reactgular

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2 Comments

Swoop Over a year ago

Yes, this is an example of what NOT to do. I just couldn't get the behavior. If the pointer is getting reset, why then wouldn't the output be just "Apple => 1" over and over?

Reactgular Over a year ago

@Swoop PHP works with arrays in a copy when modified approach. When you use $desserts[]=$pie; the values are not actually copied to $desserts until something is changed. So reset might not happen until PHP sees that $pie and $desserts are no longer the same, and that could be when $pie hits the end of the array.

qwertynl · Accepted Answer · 2013-11-22 15:49:51Z

1

You can avoid this loop by using a foreach loop instead of the while loop:

$pie = array('Apple' => 1, 'Cherry' => 3, 'Pumpkin' => 6);
foreach($pie as $k => $v){
    echo "$k => $v\n";
    $desserts[]=$pie;
}

answered Nov 22, 2013 at 15:49

qwertynl

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Comments

Bas Kuis · Accepted Answer · 2013-11-22 16:03:54Z

1

$pie = array('Apple' => 1, 'Cherry' => 3, 'Pumpkin' => 6);
reset($pie);
while (list($k, $v) = each($pie)) {
    echo "$k => $v\n";
    $desserts[]=$pie;
}

Doesn't make that much sense. You're basically duplicating the $pie array for each entry in $pie.

Using foreach would be safer. Would the sample below reflect more closely the intended result perhaps?

$pie = array('Apple' => 1, 'Cherry' => 3, 'Pumpkin' => 6);
$desserts = array();
foreach($pie as $fruit => $count){
    $dessert = array(
        'fruit' => $fruit,
        'count' => $count
    );
    $desserts[] = $dessert;
}

answered Nov 22, 2013 at 16:03

Bas Kuis

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