Values of array changing without modification, using foreach loop

Question

I have a sample code. Its strange! Even not making any changes in defined array but still my defined array's value got changed.

$myarr = array(1, 2, 3, 4);
foreach ($myarr as &$myvalue) {
    $myvalue = $myvalue * 2;
}
print_r($myarr); // Output - Array ( [0] => 2 [1] => 4 [2] => 6 [3] => 8 )

Can you please explain, How it happen?

Read about [references](http://php.net/manual/en/language.references.php). Using them in a loop produces unexpected results. This is explained in the documentation: http://php.net/manual/en/language.references.whatdo.php — axiac, Sep 09 '18 at 10:07
What do you mean? You are multiplying each value in your array by 2, in the foreach statement: $myvalue = $myvalue * 2; The output is correct. — Bandreid, Sep 09 '18 at 10:08

score 2 · Accepted Answer · answered Sep 09 '18 at 10:10

2

You do change the original array because you are using & in your array loop.

This signal for references as @axiac comments.

In order to avoid changes on your original array use the following for loop:

foreach ($myarr as $myvalue)

answered Sep 09 '18 at 10:10

dWinder

11,597
3
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Thanks David, I could understand that what exactly mean of reference. – sandeep soni Sep 09 '18 at 10:31

score 0 · Answer 2 · answered Sep 09 '18 at 10:13

0

you are passing address of myarr, where myvalue uses same address of myarr that why its changing

use this

  $myarr = array(1, 2, 3, 4);
    foreach ($myarr as $myvalue) {
        $myvalue = $myvalue * 2;
    }
    print_r($myarr);

answered Sep 09 '18 at 10:13

Vaibhav Shettar

790
2
9
20

score 0 · Answer 3 · answered Sep 09 '18 at 10:13

0

have you tried it without the "&" before $myvalue?

Please read here PHP: What does a & in front of a variable name mean?

answered Sep 09 '18 at 10:13

Alexander Voss

57
4

Values of array changing without modification, using foreach loop

3 Answers3