Why does PHP's foreach with pass-by-reference (&) change array elements unexpectedly?

PHP Pass by Reference in Foreach

In PHP, variables can be either normal variables or reference variables. When assigning a reference of a variable to another variable, the latter becomes a reference variable, essentially binding to the same memory location as the original. This behavior has implications when using the foreach loop with the pass-by-reference syntax (&).

Consider the following code:

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

foreach ($a as &$v) {

}

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

The output of this code is:

zero one two two

Why does this occur?

In the first foreach loop, each element of $a is assigned as a reference to the variable $v. This means that any modification to $v also modifies the corresponding element in $a.

Specifically, during the last iteration of the first loop, $v is assigned the value of $a[3] (i.e., 'three'). However, this assignment creates a reference relationship, so $a[3] becomes a reference variable (even though it is not assigned explicitly using &).

In the second foreach loop, the values of the elements in $a are printed. However, since $a[3] is now a reference variable, its value changes along with the value of $v. When $v is assigned the value 'two' during the third iteration, $a[3] also becomes 'two'.

As a result, in the last iteration, $v (which still points to $a[3]) has the value 'two', and 'two' is printed. This explains why 'two' is repeated in the last iteration, instead of printing 'three' as one might intuitively expect.

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