PHP array passing is value passing rather than reference passing concept correction_PHP tutorial

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’s worth noting!

Copy code The code 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:

Copy code The 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 "]

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