PHP:非匿名函数的 use() 替代方法

Question

你可以这样做:

$external = 1;
$change = function($number) use(&$external) {
    $external = $number;
};

$change(5);
echo $external; //> 5

但是你不能这样做:

$external = 1;
function change($number) use(&$external) {
    $external = $number;
}

你会得到:

Parse error: syntax error, unexpected 'use' (T_USE), expecting '{'.

什么是替代品？

Answer 1

use()仅在闭包中用于从父作用域继承变量。来自手册:

Closures may also inherit variables from the parent scope. Any such variables must be passed to the use language construct.

如果您想在常规函数中通过引用使用变量，请使用:

$external = 1;

function change($number, &$external) {
    $external = $number;
}
change(5, $external);

或者不通过它使用它作为全局(如果 $external 在全局范围内):

$external = 1;

function change($number) {
    $GLOBALS['external'] = $number;
}
change(5);