PHP Array Value problem

Go directly to the code. Hope can get the analysis result of the great God.

$test = array(1, 2, 3, 4, 5);foreach ($test as &$value) {    $value = 0;}unset($value);print_r($test);// 结果如下Array(    [0] => 0    [1] => 0    [2] => 0    [3] => 0    [4] => 0)$test = array(1, 2, 3, 4, 5);foreach ((array)$test as &$value) {    $value = 0;}unset($value);print_r($test);// 结果如下Array(    [0] => 1    [1] => 2    [2] => 3    [3] => 4    [4] => 5)

As shown in the code above, just using the array coercion type conversion, the result is very different, I want to know why?

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Go directly to the code. Hope can get the analysis result of the great God.

$test = array(1, 2, 3, 4, 5);foreach ($test as &$value) {    $value = 0;}unset($value);print_r($test);// 结果如下Array(    [0] => 0    [1] => 0    [2] => 0    [3] => 0    [4] => 0)$test = array(1, 2, 3, 4, 5);foreach ((array)$test as &$value) {    $value = 0;}unset($value);print_r($test);// 结果如下Array(    [0] => 1    [1] => 2    [2] => 3    [3] => 4    [4] => 5)

As shown in the code above, just using the array coercion type conversion, the result is very different, I want to know why?

Let's look at this example.

$test = array(1, 2, 3, 4, 5);$test1 = $test;foreach ($test1 as &$value) {    $value = 0;}print_r($test);Array(    [0] => 1    [1] => 2    [2] => 3    [3] => 4    [4] => 5)

Look at this example again.

$test = array(1, 2, 3, 4, 5);$test1 = &$test;foreach ($test1 as &$value) {    $value = 0;}print_r($test);Array(    [0] => 0    [1] => 0    [2] => 0    [3] => 0    [4] => 0)

Above even an example is the first assignment, then the foreach, but the first is a value, and the second is a reference (address), the value of the original array to do a copy assignment to the new variable, they point in memory is not the same place.

In the second example of your (array) $test, this is actually the equivalent of creating a new array, this array and test must not point to the same memory address, so the assignment operation in foreach does not modify the value in the original test variable

(array) $test this itself is an assignment operation equal to $a = (array) $test and then the foreach $a

The assignment of arrays (array) always involves a copy of the value. Copy an array using the reference operator & by reference

php$test = array(1, 2, 3, 4, 5);$test2 = (array)$test;$test2[0] = 0;print_r($test);//[1,2,3,4,5]print_r($test2);//[0,2,3,4,5]

When a foreach iterates through an array, he will release value for every traversal, and if you add a reference, he will not release it and be overwritten by the subsequent one. Believe me,

Two kinds of code i output opcode to discuss for opcode.
When doing such operations as foreach (xxx as XX), the invoked instruction is fe_rest to reset the array's iterator.
From my red box can be clearly seen, the first time to reset the use of the! 0, that is, $test, the second use of a, that is (array) $test.
So the first time you change, you change the contents of the $test, and the second time, the change is (array) $test the contents of this variable, meaning that the second time is not changed to $test

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