这里有一个清单来确定你是否违反了利斯科夫法则。

如果你违反了以下项目之一->，你违反了里斯科夫。 如果你不违反任何->不能得出任何结论。

检查表:

No new exceptions should be thrown in derived class: If your base class threw ArgumentNullException then your sub classes were only allowed to throw exceptions of type ArgumentNullException or any exceptions derived from ArgumentNullException. Throwing IndexOutOfRangeException is a violation of Liskov. Pre-conditions cannot be strengthened: Assume your base class works with a member int. Now your sub-type requires that int to be positive. This is strengthened pre-conditions, and now any code that worked perfectly fine before with negative ints is broken. Post-conditions cannot be weakened: Assume your base class required all connections to the database should be closed before the method returned. In your sub-class you overrode that method and left the connection open for further reuse. You have weakened the post-conditions of that method. Invariants must be preserved: The most difficult and painful constraint to fulfill. Invariants are sometimes hidden in the base class and the only way to reveal them is to read the code of the base class. Basically you have to be sure when you override a method anything unchangeable must remain unchanged after your overridden method is executed. The best thing I can think of is to enforce these invariant constraints in the base class but that would not be easy. History Constraint: When overriding a method you are not allowed to modify an unmodifiable property in the base class. Take a look at these code and you can see Name is defined to be unmodifiable (private set) but SubType introduces new method that allows modifying it (through reflection): public class SuperType { public string Name { get; private set; } public SuperType(string name, int age) { Name = name; Age = age; } } public class SubType : SuperType { public void ChangeName(string newName) { var propertyType = base.GetType().GetProperty("Name").SetValue(this, newName); } }

还有2项:方法参数的逆变性和返回类型的协方差。但这在c#中是不可能的(我是c#开发人员)，所以我不关心它们。

A square is a rectangle where the width equals the height. If the square sets two different sizes for the width and height it violates the square invariant. This is worked around by introducing side effects. But if the rectangle had a setSize(height, width) with precondition 0 < height and 0 < width. The derived subtype method requires height == width; a stronger precondition (and that violates lsp). This shows that though square is a rectangle it is not a valid subtype because the precondition is strengthened. The work around (in general a bad thing) cause a side effect and this weakens the post condition (which violates lsp). setWidth on the base has post condition 0 < width. The derived weakens it with height == width.

因此，可调整大小的正方形不是可调整大小的矩形。

一些补充:我想知道为什么没有人写基类的不变量、前提条件和后置条件，这些派生类必须遵守。 对于派生类D来说，基类B完全可转换，类D必须服从某些条件:

基类的内变体必须由派生类保留 派生类不能加强基类的先决条件 派生类不能削弱基类的后置条件。

因此派生类必须知道基类施加的上述三个条件。因此，子类型的规则是预先确定的。这意味着，只有当子类型遵守某些规则时，才应该遵守'IS A'关系。这些规则，以不变量、前置条件和后置条件的形式，应该由正式的“设计契约”来决定。

关于这个问题的进一步讨论可以在我的博客:利斯科夫替换原理

长话短说，让我们留下矩形矩形和正方形，实际的例子，当扩展一个父类时，你必须要么保留确切的父API，要么扩展IT。

假设您有一个基本ItemsRepository。

class ItemsRepository
{
    /**
    * @return int Returns number of deleted rows
    */
    public function delete()
    {
        // perform a delete query
        $numberOfDeletedRows = 10;

        return $numberOfDeletedRows;
    }
}

以及扩展它的子类:

class BadlyExtendedItemsRepository extends ItemsRepository
{
    /**
     * @return void Was suppose to return an INT like parent, but did not, breaks LSP
     */
    public function delete()
    {
        // perform a delete query
        $numberOfDeletedRows = 10;

        // we broke the behaviour of the parent class
        return;
    }
}

然后，您可以让客户端使用Base ItemsRepository API并依赖它。

/**
 * Class ItemsService is a client for public ItemsRepository "API" (the public delete method).
 *
 * Technically, I am able to pass into a constructor a sub-class of the ItemsRepository
 * but if the sub-class won't abide the base class API, the client will get broken.
 */
class ItemsService
{
    /**
     * @var ItemsRepository
     */
    private $itemsRepository;

    /**
     * @param ItemsRepository $itemsRepository
     */
    public function __construct(ItemsRepository $itemsRepository)
    {
        $this->itemsRepository = $itemsRepository;
    }

    /**
     * !!! Notice how this is suppose to return an int. My clients expect it based on the
     * ItemsRepository API in the constructor !!!
     *
     * @return int
     */
    public function delete()
    {
        return $this->itemsRepository->delete();
    }
} 

当用子类替换父类破坏了API的契约时，LSP就被破坏了。

class ItemsController
{
    /**
     * Valid delete action when using the base class.
     */
    public function validDeleteAction()
    {
        $itemsService = new ItemsService(new ItemsRepository());
        $numberOfDeletedItems = $itemsService->delete();

        // $numberOfDeletedItems is an INT :)
    }

    /**
     * Invalid delete action when using a subclass.
     */
    public function brokenDeleteAction()
    {
        $itemsService = new ItemsService(new BadlyExtendedItemsRepository());
        $numberOfDeletedItems = $itemsService->delete();

        // $numberOfDeletedItems is a NULL :(
    }
}

你可以在我的课程中学习更多关于编写可维护软件的知识:https://www.udemy.com/enterprise-php/

以Board数组的形式实现ThreeDBoard会有用吗?

也许你想把不同平面上的ThreeDBoard切片作为一个板。在这种情况下，您可能希望为Board抽象出一个接口(或抽象类)，以允许多种实现。

就外部接口而言，您可能希望为TwoDBoard和ThreeDBoard提取一个Board接口(尽管上述方法都不适合)。

使用LSP的一个重要例子是在软件测试中。

如果我有一个类a，它是B的一个符合lsp的子类，那么我可以重用B的测试套件来测试a。

为了完全测试子类A，我可能需要添加更多的测试用例，但至少我可以重用所有超类B的测试用例。

实现这一点的一种方法是构建McGregor所说的“用于测试的并行层次结构”:我的ATest类将继承BTest。然后需要某种形式的注入来确保测试用例使用类型A的对象而不是类型B的对象(一个简单的模板方法模式就可以了)。

注意，对所有子类实现重用超级测试套件实际上是一种测试这些子类实现是否与lsp兼容的方法。因此，人们也可以主张应该在任何子类的上下文中运行超类测试套件。

另请参阅对Stackoverflow问题的回答“我是否可以实现一系列可重用测试来测试接口的实现?”