Need a help to debug PHP code

Question

I have following code as below,

 <?php
    $a = [1,2,3];
    foreach($a as &$val) {
        $val = $val + 1;
    }

    foreach($a as $val) {
        $val = $val - 1;
    }

    var_dump($a);

    // output 2,3,1
?>

I got output 2,3,1 as final array instead of 2,3,4 and i can't understand how php is interpreting this code, Can anyone help me to understand how the things is going here?

@AlonEitan No, the first pass-by-reference should create 2,3,4 and be unaffected by the seconded foreach. — BenM
– BenM, Commented May 12, 2018 at 17:48
Accessing the loop variable of foreach by reference means searching for trouble. Don't do it! — axiac
– axiac, Commented May 12, 2018 at 20:09
The answer to your question is explained in the documentation of foreach in a big red warning box. — axiac
– axiac, Commented May 12, 2018 at 20:11

BenM · Accepted Answer · 2018-05-18 13:55:15Z

5

You need to call unset() on the reference in your first foreach() to get the expected behaviour:

$a = [1, 2, 3];

foreach($a as &$val) 
{
    $val = $val + 1;
}

unset($val);

// $a = [2, 3, 4];

See the note in the documentation for this:

Reference of a $value and the last array element remain even after the foreach loop. It is recommended to destroy it by unset().

edited May 18, 2018 at 13:55

answered May 12, 2018 at 17:51

BenM

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6 Comments

Ashish Rana Over a year ago

I think question is about why output is [2,3,1] instead of [2,3,4]

BenM Over a year ago

without an unset($value), $value is still a reference to the last item: $arr[3]. It's in the docs, guys.

rob006 Over a year ago

unset() should be called after foreach, not inside of it.

BenM Over a year ago

@rob006 It's actually irrelevant in this instance: eval.in/1006699

rob006 Over a year ago

@BenM result may be the same, but it does not make any sense to put it inside of foreach, it only adds unnecessary overhead.

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