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，现在我陷入了静态匿名函数"中.

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

我在教程中找到了此内容( 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)

对象方向

• Lambda函数之所以是闭包，是因为它们自动绑定到创建它们的类的范围.
范围中并不总是需要
• '$this'.
• 删除'$this'可以节省内存.
• 您可以通过将Lambda函数声明为静态来阻止此行为."

• 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."

此代码有什么问题?

我收到此错误:

解析错误:解析错误，期望在第11行的C:\ wamp \ www \ z-final \ a.php中出现"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返回静态函数(){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，在5.4+中是完全有效的语法.

Yes, that is perfectly valid syntax in 5.4+.

基本上，它阻止了当前类与闭包的自动绑定(实际上，它阻止了所有绑定，但稍后会进行更多介绍).

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，则闭包bar()返回将把$this设置为null.就像您对它进行了静态调用一样.但是baz()会将$this设置为您在其上调用过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:未定义的变量:第5行的/in/Bpd3d中的变量

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

NULL

还有

$baz = $f->baz();

$baz();

结果

object(Foo)#1(0){

object(Foo)#1 (0) {

}

有道理吗?很好.

现在，如果我们在函数外部定义闭包，会发生什么情况

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

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

我们收到null(和通知)

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

正如您所期望的那样，我们得到StdClass

We get StdClass, just as you'd expect

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

我们收到null(和通知)

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

这是使事情变得有趣的地方.现在，我们得到警告，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:未定义的变量:第9行的/in/h63iF中的变量

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

NULL

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绑定$this，也永远不会将对象绑定到它...

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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