PHP - array_push in foreach loop of the same array doesn't update the array

Question

I need to be able to echo 1 and 2 with an array_push in the same foreach loop of that array.

<?php
$arr = array('1');

foreach ($arr as &$arr_number) {
   echo "$arr_number\n"; //this print 1
   array_push($arr, '2');
}

foreach ($arr as &$arr_number) {
   echo "$arr_number\n"; //this print 1 and then 2
}
?>

EDIT: Solved it myself by not using a foreach but a while loop

    <?php
    $arr = array('1');

 while ( list($key, $value) = each($arr) ) {
    echo "$value\n";

    if ( !in_array('2', $arr) ) {
       array_push($arr, '2');
    }
 }
 ?>

@zerkms: yeah, probably no point in using references... doesn't matter, code still doesn't work with foreach without references — Zurd
– Zurd, Commented Mar 4, 2014 at 7:42

Shankar Narayana Damodaran · Accepted Answer · 2014-03-04 07:30:50Z

3

I need to be able to echo 1 and 2 with an array_push in the same foreach loop of that array.

Implode it.

<?php
$arr = array('1');

foreach ($arr as &$arr_number) {

   array_push($arr, '2');
   echo implode(' ',$arr); //"prints" 1 2
}

answered Mar 4, 2014 at 7:30

Shankar Narayana Damodaran

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10 Comments

KarelG Over a year ago

would that not be an infinite loop ?

KarelG Over a year ago

you're pushing an variable into an array which you're looping through. I thought that the iterator would detect the newly added item and thus continuing the loop. But it's an instance at that moment. My excuses for my confusion.

AyB Over a year ago

@KarelG The implode function above simply concatenates all the array values together with a space. It won't affect the loop. Also the iterator does not detect the newly added value which is why the OP is receiving only 1 as output.

Hereblur Over a year ago

@KarelG No, it's not. see the first answer of stackoverflow.com/questions/2348077/…

Zurd Over a year ago

@KarelG yes that would be an infinite loop, I had to kill PHP :) I added the check in_array to make it work

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