Why cant i push element onto sub array in foreach loop?

Question

I'm trying to understand something about arrays in for each loops that might be obvious to many.

When i loop through my multi-dimensional array, i attempt to find sub arrays with no third element. If they have no third element, i want to add a third element to that sub array with a specific value.

$testArray = array (
    array("Green", "Yellow", "Blue"),
    array("Brown", "Silver"),
    array("Orange", "Pink", "Black"),
);

When i use the foreach loop:

foreach ( $testArray as $key => $array ) {
    if (count($array) == '2') {
        $array[] = "None";
    };
}

No errors are thrown but nothing happens. When i use the for each loop:

foreach ( $testArray as $key => $array ) {
    if (count($array) == '2') {
        $testArray[$key][] = "None";
    };
}

It works as expected.

Sorry for the long preamble, my questions is:

Why aren't these two foreach loops doing the same thing? Thanks!

Answer 1

The problem here lies in the fact that foreach iterates over iterables and sets the iteration variable by value. This means that the $array which you are dealing with in the foreach is not the same value of the $testArray.

To rememdy this (and avoid introducing an $index variable to mutate an item in the array), you will need to tell foreach to pass the value by reference. References are PHP's answer to C-style pointers. If a variable references another, both variables point to the same value, so a modification to the contents of one is in effect a modification to the other. In your foreach, you can use &$array to have the loop pass you the items of $testArray by reference instead of by value:

foreach ( $testArray as $key => &$array ) {
    if (count($array) == '2') {
        $array[] = "None";
    }
}

(Codepad Demo)

This aligns with PHP's references, where one variable can be made to reference another like so:

$a = array(1, 2, 3);
$b = &$a;
$b[] = 4;

print_r($a); // 1, 2, 3, 4

(Codepad Demo)

You experience a similar phenomenon with functions:

function byValue($a) {
    $a[] = 4;
}

function byRef(&$a) {
    $a[] = 5;
}

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

byValue($a);
print_r($a); // 1, 2, 3

byRef($a);
print_r($a); // 1, 2, 3, 5

(Codepad Demo)

The references section of the PHP docs has some examples about this syntax of foreach. Also note this (somewhat) related, but interesting read on foreach and references.

---

Also, on an unrelated note if you weren't aware: you don't need a semicolon after closing a block with } in PHP unless you're doing something like assigning a closure to a variable:

$someFunc = function($a) { return $a; }; //need one here

if(1 + 2 == 4) {
   echo "I just broke math";
} // but you don't need one here

Answer 2

Because you need to access $testArray entries "by reference"

foreach ( $testArray as &$array ) {
    if (count($array) == 2) {
        $array[] = "None";
    };
}
unset($array);

Answer 3

foreach loops do not pass the elements by reference. To get the first array to do what you want, it would have to be:

foreach ( $testArray as $key => &$array ) {
    if (count($array) == '2') {
        $array[] = "None";
    };
}