typehinting:方法应接受任何作为对象的$ arg [英] typehinting: method should accept any $arg that is an object

问题描述

我有一个'Collection'类，它具有一个add方法. add方法应仅接受对象.因此，这是所需的行为:

I have a class 'Collection', which has an add method. The add method should only accept objects. So this is the desired behaviour:

$x=5;//arbitrary non-object
$obj=new Foo; //arbitrary object

$collection=new Collection;
$collection->add($obj); //should be acceptable arg, no matter the actual class
$collection->add($x); //should throw an error because $x is not an object

根据PHP手册，可以通过在$arg前面加上类名来键入提示方法.由于所有PHP类都是stdClass的子类，因此我认为此方法签名可以工作:

According to the PHP manual, one can typehint methods by prefacing the $arg with a class name. Since all PHP classes are children of stdClass, I figured this method signature would work:

public function add(stdClass $obj);

但是它失败并显示参数必须是stdClass的实例".

But it fails with "Argument must be an instance of stdClass".

如果我将签名更改为我定义的父类，那么它将起作用:

If I change the signature to a parent class defined by me, then it works:

class Collection {
  public function add(Base $obj){
    //do stuff
  }
}

$collection->add($foo); //$foo is class Foo which is an extension of Base

有人知道如何为通用对象键入提示吗?

Does anyone know how to type hint for a generic object?

推荐答案

与Java的Object类不同，PHP 没有对象的基类.对象不继承stdClass:这是默认的对象实现，而不是基类.因此，不幸的是，您不能为PHP中的所有对象键入提示.您必须执行以下操作:

Unlike Java's Object class, PHP does not have a base class for objects. Objects do not inherit stdClass: it's a default object implementation, not a base class. So, unfortunately, you can't type hint for all objects in PHP. You have to do something like:

class MyClass {
    public function myFunc($object) {
        if (!is_object($object))
             throw new InvalidArgumentException(__CLASS__.'::'.__METHOD__.' expects parameter 1 to be object");
    }
}

幸运的是，PHP已经为此目的定义了InvalidArgumentException类.

Luckily, PHP already defines the InvalidArgumentException class for that purpose.

这篇关于typehinting:方法应接受任何作为对象的$ arg的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！