PHP 静态匿名函数真的有效吗? [英] Does a PHP static anonymous function really work?

问题描述

我正在尝试学习 PHP，但现在我被困在静态匿名函数"中.

I'm trying to learn PHP, and now I'm stuck in 'static anonymous function'.

我在教程中找到了这个(http://www.slideshare.net/melechi/php-53-part-2-lambda-functions-closures-presentation)

I found this in a tutorial (http://www.slideshare.net/melechi/php-53-part-2-lambda-functions-closures-presentation)

面向对象

• Lambda 函数是闭包，因为它们会自动绑定到创建它们的类的作用域.
• '$this' 在作用域中并不总是需要的.
• 删除$this"可以节省内存.
• 您可以通过将 Lambda 函数声明为静态来阻止这种行为."

• Lambda Functions are Closures because they automatically get bound to the scope of the class that they are created in.
• '$this' is not always needed in the scope.
• Removing '$this' can save on memory.
• You can block this behaviour by declaring the Lambda Function as static."

这段代码有什么问题?

我收到此错误:

解析错误:解析错误，在第 11 行的 C:\wamp\www\z-final\a.php 中期待`T_PAAMAYIM_NEKUDOTAYIM'

Parse error: parse error, expecting `T_PAAMAYIM_NEKUDOTAYIM' in C:\wamp\www\z-final\a.php on line 11

为什么此代码行不起作用return static function(){var_dump($this);};"?

Why this code line doesn't work "return static function(){var_dump($this);};" ?

class foo
{
    public function getLambda()
    {
        return function(){var_dump($this);};
    }

    public function getStaticLambda()
    {
        return static function(){var_dump($this);};
    }
}

$foo = new foo();
$lambda = $foo->getLambda();
$staticLambda = $foo->getStaticLambda();
$lambda();
$staticLambda();

推荐答案

是的，这是 5.4+ 中完全有效的语法.

Yes, that is perfectly valid syntax in 5.4+.

基本上，它可以防止当前类自动绑定到闭包(实际上，它可以防止所有绑定，但稍后会详细介绍).

Basically, it prevents auto-binding of the current class to the closure (in fact, it prevents all binding, but more on that later).

class Foo {
    public function bar() {
        return static function() { var_dump($this); };
    }
    public function baz() {
        return function() { var_dump($this); };
    }
}

如果我们在 5.4+ 上实例化它，闭包 bar() 返回的 $this 将设置为 null.就像您对它进行了静态调用一样.但是 baz() 会将 $this 设置为您调用 baz() 的 foo 实例.

If we instantiate that on 5.4+, the closure bar() returns will have $this set to null. Just as if you made a static call to it. But baz() would have $this set to the foo instance you called baz() on.

所以:

$bar = $f->bar();

$bar();

结果:

注意:未定义变量:this in/in/Bpd3d on line 5

Notice: Undefined variable: this in /in/Bpd3d on line 5

空

和

$baz = $f->baz();

$baz();

结果

object(Foo)#1 (0) {

object(Foo)#1 (0) {

}

有意义吗?太好了.

现在，如果我们采用在函数外部定义的闭包会发生什么:

Now, what happens if we take closures defined outside of a function:

$a = function() { var_dump($this); };
$a();

我们得到 null(和一个通知)

We get null (and a notice)

$c = $a->bindTo(new StdClass());
$c();

我们得到 StdClass，正如你所期望的

We get StdClass, just as you'd expect

$b = static function() { var_dump($this); };
$b();

我们得到 null(和一个通知)

We get null (and a notice)

$d = $b->bindTo(new StdClass());
$d();

这就是事情变得有趣的地方.现在，我们得到一个警告、一个通知和 null:

This is where things get interesting. Now, we get a warning, a notice, and null:

警告:无法在第 12 行的/in/h63iF 中将实例绑定到静态闭包

Warning: Cannot bind an instance to a static closure in /in/h63iF on line 12

注意:未定义变量:this in/in/h63iF on line 9

Notice: Undefined variable: this in /in/h63iF on line 9

空

因此在 5.4+ 中，您可以声明一个静态闭包，这将导致它永远不会将 $this 绑定到它，也永远不能将一个对象绑定到它...

So in 5.4+, you can declare a static closure, which results in it never getting $this bound to it, nor can you ever bind an object to it...

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