Php function to return array: why I have to pass by reference?

Question

I'm coming to Php from Javascript and I can't understand why, to get, transform and return an array in Php, you have to use the "&" aka "reference" symbol in the foreach()...

       $some_array = array(1, 2, 3);



       function test($a) {
           foreach ($a as $k => &$v) {
               $v = 'Hello!';
           };
           return $a;
       };

       $result = test($some_array);


       var_dump($result); // array('Hello!', 'Hello!', 'Hello!');

But with:

 function test($a) {
           foreach ($a as $k => $v) {
               $v = 'Hello!';
           };
           return $a;
       };

[...]

var_dump($result); // array(1, 2, 3)

Another way, which may be easier to use for complex arrays is to set the original parameter array with the new value. You have the key of the current element. It is safe to do this in a foreach. see: eval.in/726978. i.e. it gives the same result without references. You need to know the element keys. Just a different way of getting the same effect. — Ryan Vincent
– Ryan Vincent, Commented Jan 30, 2017 at 15:46

Community · Accepted Answer · 2020-06-20 09:12:55Z

3

To understand what "passed by value" and "passed by reference" are, you should read the docs:

By default, function arguments are passed by value (so that if the value of the argument within the function is changed, it does not get changed outside of the function). To allow a function to modify its arguments, they must be passed by reference.

To have an argument to a function always passed by reference, prepend an ampersand (&) to the argument name in the function definition.

The docs on foreach construct also states:

In order to be able to directly modify array elements within the loop precede $value with &. In that case the value will be assigned by reference.

You may choose to pass by reference and change the value directly (&$v), or you may choose to pass by value, but in this case you have to change the array value using the element key in $k:

foreach ($a as $k => $v)
{
    $a[$k] = 'Hello!';
};

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Jan 30, 2017 at 12:55

Kevin Kopf

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4 Comments

RiggsFolly Over a year ago

Actually its nothing to do with the function argument in this case. It's purely to do with the reference used (or not used) on the foreach

Kevin Kopf Over a year ago

@RiggsFolly dam right, I fixed my answer a bit, thanx

RiggsFolly Over a year ago

I dont really understand your last sentence either, I am not sure what you are saying there

Kevin Kopf Over a year ago

@RiggsFolly is this better? :)

RiggsFolly · Accepted Answer · 2017-01-30 13:13:58Z

1

If you just pass the array to the foreach without using the reference &, you are passing a copy of the array into the foreach.

When you then make changes to the array (the copy) you make those changes to the copy. The copy of course is thrown away once the foreach finishes.

So when you return $a after the foreach has completed, you are actually returning the copy that was passed to the function as the parameter, which is unchanged.

Even if you were to pass the array, by reference, to the function it would make no difference as it is the foreach that either makes a disposable copy, or uses the original array, depending upon the use of the & reference instruction in the actual foreach code line.

edited Jan 30, 2017 at 13:13

answered Jan 30, 2017 at 12:55

RiggsFolly

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