Is PHP array passing by reference?

PHP array transfer is not by reference, but by value; when calling a function, assigning the PHP array as an actual parameter to the formal parameter and modifying it in the function will not affect the array itself, explaining this process The transfer in is by value, and the array variable is not a reference to the array itself.

The operating environment of this article: Windows 10 system, PHP version 8.1, Dell G3 computer

Is php array passing by reference?

##Array passing in PHP is by value rather than by reference.

When calling a function, assign the PHP array as an actual parameter to the formal parameter, and modifying it in the function will not affect the array itself.

Explain that the transfer in this process is by value. The array variable is not a reference to the array itself. The PHP array itself exists in the form of a value, and the formal parameter is a copy of the array.

This is very different from other languages ​​​​(such as c, Js, etc.), so it is worth noting!

The example is as follows:

$arr = array(
    'name' => 'corn',
    'age' => '24',
);
test_arr($arr);
function test_arr($arr){
    $arr['name'] = 'qqyumidi';
}
print_r($arr); //result: Array ( [name] => corn [age] => 24 )

js code is as follows:

var arr = new Array('corn', '24');
test_arr(arr);
function test_arr(arr){
    arr[0] = 'qqyumidi';
}
console.log(arr);  //result:["qqyumidi", "24"]

If you need to use the reference passing effect for value passing in PHP, you can add in front of the formal parameter Take the address character &.

$aa = 100;
test_vars($aa);
function test_vars(&$aa){
    $aa = 200;
}
print_r($aa);   //result: 200

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