Why Does My PHP `foreach` Loop Produce Unexpected Results When Using Pass-by-Reference?

Passing by Reference in foreach: Understanding the Unexpected Behavior

The PHP foreach loop iterates over each element in an array, but in some cases, its output can be surprising. Consider the following code:

$a = ['zero', 'one', 'two', 'three'];

foreach ($a as &$v) {

}

foreach ($a as $v) {
  echo $v . PHP_EOL;
}

Expectedly, the output should be 'zero', 'one', 'two', and 'three'. However, if you run this code, you'll get 'zero', 'one', 'two', 'two' instead.

This unexpected behavior stems from PHP's concept of variables by reference. In the first foreach loop, we use the reference operator "&" to create a reference variable ($v) to each element in the $a array. This means that any changes made to $v will also affect the corresponding element in $a.

Initially, all elements in $a are normal variables. However, after the first loop, only $a[3] remains a reference variable. This is because while iterating through the array, the previous reference variable ($v) is overwritten with each new element.

In the second loop, when we echo $v, it points to $a[3], which was set to 'two' in the previous iteration. Therefore, 'two' is echoed instead of 'three' in the last iteration.

By understanding the difference between normal variables and reference variables, and how they interact in a foreach loop, we can avoid such unexpected behaviors and ensure that our code operates as intended.

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