在 PHP 中将实例方法作为参数传递
我想创建一个监听器类
class Listener {
var $listeners = array();
public function add(callable $function) {
$this->listeners[] = $function;
}
public function fire() {
foreach($this->listeners as $function) {
call_user_func($function);
}
}
}
class Foo {
public function __construct($listener) {
$listener->add($this->bar);
}
public function bar() {
echo 'bar';
}
}
$listener = new Listener();
$foo = new Foo($listener);
但是此代码失败并出现此错误:
But this code fails with this error:
注意:未定义的属性:18 行 index.php 中的 Foo::$bar
Notice: Undefined property: Foo::$bar in index.php on line 18
可捕获的致命错误:传递给 Listener::add() 的参数 1 必须是可调用的,给定 null,在第 18 行在 index.php 中调用并在第 5 行定义了 index.php
Catchable fatal error: Argument 1 passed to Listener::add() must be callable, null given, called in index.php on line 18 and defined index.php on line 5
我做错了什么?
推荐答案
在 PHP 5.4 之前,没有名为
callable
的类型,所以如果将它用作类型提示,则表示名为callable
的类".如果您使用 PHP >= 5.4,callable
是一个有效的提示.Before PHP 5.4, there was no type named
callable
, so if you use it as a type hint, it means "the class namedcallable
". If you use PHP >= 5.4,callable
is a valid hint.可调用对象由描述可调用对象名称的字符串(例如函数名或类方法名)或数组指定,其中第一个元素是对象的实例,第二个元素是要调用的方法的名称.
A callable is specified by a string describing the name of the callable (a function name or a class method name for example) or an array where the first element is an instance of an object and the second element is the name of the method to be called.
对于 PHP <5.4、替换
For PHP < 5.4, replace
public function add(callable $function)
与:
public function add($function)
调用它:
$listener->add(array($this, 'bar'));
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