Why Does the Last Element Duplicate in PHP's Foreach Loop with Pass-by-Reference?

PHP Foreach Pass by Reference Enigma: Mysterious Last Element Duplication

In PHP, when you employ the foreach loop with a pass-by-reference assignment (e.g., foreach ($arr as &$item)), unexpected behavior can arise. Consider this perplexing example:

$arr = ["foo", "bar", "baz"];

foreach ($arr as &$item) { /* do nothing by reference */ }
print_r($arr);

foreach ($arr as $item) { /* do nothing by value */ }
print_r($arr); // $arr has inexplicably changed

This code outputs:

Array
(
    [0] => foo
    [1] => bar
    [2] => baz
)
Array
(
    [0] => foo
    [1] => bar
    [2] => bar
)

Explaining the Duplication

After the first foreach loop, each element of $arr is still referenced by $item. When the second loop iterates, it replaces the value of each element with the value of $item, which happens to be the last element of the array. This means that each element of $arr is set to the value of $arr[2], leading to the duplication of the last element in the output.

Debugging the Output

To illustrate this behavior, let's debug the output by adding print statements to each foreach iteration:

foreach ($arr as &$item) {
    echo "Item: $item<br>";
    print_r($arr);
    echo "<br>";
}

foreach ($arr as $item) {
    echo "Item: $item<br>";
    print_r($arr);
    echo "<br>";
}

This outputs:

Item: foo
Array ( [0] => foo [1] => bar [2] => baz )

Item: bar
Array ( [0] => foo [1] => bar [2] => baz )

Item: baz
Array ( [0] => foo [1] => bar [2] => baz )

Item: foo
Array ( [0] => foo [1] => bar [2] => foo )

Item: bar
Array ( [0] => foo [1] => bar [2] => bar )

Item: bar
Array ( [0] => foo [1] => bar [2] => bar )

You can clearly see that each element of $arr is set to the value of $item, which changes to the last element of the array during the second foreach loop.

Bug or Intended Behavior?

This behavior is not a bug. It is a consequence of passing by reference. The foreach loop simply assigns the value of the current element to the variable specified in the loop header. In this case, by referencing $item, we are modifying the original array elements in the second loop. This is equivalent to the following code:

for ($i = 0; $i <p>Therefore, the observed behavior is not a bug but a result of the intended semantics of pass-by-reference in PHP. To avoid such behavior, use pass-by-value in the second foreach loop by simply assigning the value of each element to $item: foreach ($arr as $item).</p>

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