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。

使用指向基类的指针或引用的函数必须能够在不知道它的情况下使用派生类的对象。

当我第一次阅读LSP时，我认为这是一个非常严格的含义，本质上等同于接口实现和类型安全强制转换。这意味着语言本身要么保证LSP，要么不保证LSP。例如，在严格意义上，ThreeDBoard当然可以取代Board，就编译器而言。

在阅读了更多关于LSP的概念之后，我发现LSP的解释通常比这更广泛。

简而言之，对于客户端代码来说，“知道”指针后面的对象是派生类型而不是指针类型的含义并不仅限于类型安全。对LSP的遵守也可以通过探测对象的实际行为进行测试。也就是说，检查对象的状态和方法参数对方法调用结果或从对象抛出的异常类型的影响。

再次回到示例，理论上Board方法可以在ThreeDBoard上很好地工作。然而，在实践中，在不妨碍ThreeDBoard打算添加的功能的情况下，防止客户端可能无法正确处理的行为差异是非常困难的。

掌握了这些知识后，评估LSP粘附性可以成为一个很好的工具，可以确定何时组合机制更适合扩展现有功能，而不是继承。

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?”

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)。

罗伯特·马丁有一篇关于利斯科夫替换原理的优秀论文。它讨论了可能违反原则的微妙和不那么微妙的方式。

论文的一些相关部分(注意，第二个例子被大量压缩):

A Simple Example of a Violation of LSP One of the most glaring violations of this principle is the use of C++ Run-Time Type Information (RTTI) to select a function based upon the type of an object. i.e.: void DrawShape(const Shape& s) { if (typeid(s) == typeid(Square)) DrawSquare(static_cast<Square&>(s)); else if (typeid(s) == typeid(Circle)) DrawCircle(static_cast<Circle&>(s)); } Clearly the DrawShape function is badly formed. It must know about every possible derivative of the Shape class, and it must be changed whenever new derivatives of Shape are created. Indeed, many view the structure of this function as anathema to Object Oriented Design. Square and Rectangle, a More Subtle Violation. However, there are other, far more subtle, ways of violating the LSP. Consider an application which uses the Rectangle class as described below: class Rectangle { public: void SetWidth(double w) {itsWidth=w;} void SetHeight(double h) {itsHeight=w;} double GetHeight() const {return itsHeight;} double GetWidth() const {return itsWidth;} private: double itsWidth; double itsHeight; }; [...] Imagine that one day the users demand the ability to manipulate squares in addition to rectangles. [...] Clearly, a square is a rectangle for all normal intents and purposes. Since the ISA relationship holds, it is logical to model the Square class as being derived from Rectangle. [...] Square will inherit the SetWidth and SetHeight functions. These functions are utterly inappropriate for a Square, since the width and height of a square are identical. This should be a significant clue that there is a problem with the design. However, there is a way to sidestep the problem. We could override SetWidth and SetHeight [...] But consider the following function: void f(Rectangle& r) { r.SetWidth(32); // calls Rectangle::SetWidth } If we pass a reference to a Square object into this function, the Square object will be corrupted because the height won’t be changed. This is a clear violation of LSP. The function does not work for derivatives of its arguments. [...]

LSP关注不变量。

经典示例由以下伪代码声明给出(实现略):

class Rectangle {
    int getHeight()
    void setHeight(int value) {
        postcondition: width didn’t change
    }
    int getWidth()
    void setWidth(int value) {
        postcondition: height didn’t change
    }
}

class Square extends Rectangle { }

现在我们有一个问题，尽管接口匹配。原因是我们违反了源自正方形和矩形数学定义的不变量。getter和setter的工作方式，矩形应该满足以下不变量:

void invariant(Rectangle r) {
    r.setHeight(200)
    r.setWidth(100)
    assert(r.getHeight() == 200 and r.getWidth() == 100)
}

然而，Square的正确实现必须违反这个不变量(以及显式后置条件)，因此它不是Rectangle的有效替代品。