PHP - 接口继承 - 声明必须兼容 [英] PHP - Interface inheritance - declaration must be compatible

问题描述

我有界面：

interface AbstractMapper
{
    public function objectToArray(ActiveRecordBase $object);
}

和班级：

class ActiveRecordBase
{
   ...
}

class Product extends ActiveRecordBase
{
   ...
}

========

但我不能这样做：

interface ExactMapper implements AbstractMapper
{
    public function objectToArray(Product $object);
}

或者这个：

interface ExactMapper extends AbstractMapper
{
    public function objectToArray(Product $object);
}

我收到错误声明必须兼容

那么，有没有办法在PHP中执行此操作？

So, is there a way to do this in PHP?

推荐答案

不，必须实现完全的接口。如果将实现限制为更具体的子类，则它不是相同的接口/签名。 PHP没有泛型或类似机制。

No, an interface must be implemented exactly. If you restrict the implementation to a more specific subclass, it's not the same interface/signature. PHP doesn't have generics or similar mechanisms.

您总是可以手动签入代码，当然：

You can always manually check in code, of course:

if (!($object instanceof Product)) {
    throw new InvalidArgumentException;
}

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