以下是这篇文章的摘录，很好地澄清了事情:

(. .为了理解一些原则，重要的是要意识到它什么时候被违反了。这就是我现在要做的。

违反这一原则意味着什么?它意味着对象不履行用接口表示的抽象所施加的契约。换句话说，这意味着您错误地识别了抽象。

考虑下面的例子:

interface Account
{
    /**
     * Withdraw $money amount from this account.
     *
     * @param Money $money
     * @return mixed
     */
    public function withdraw(Money $money);
}
class DefaultAccount implements Account
{
    private $balance;
    public function withdraw(Money $money)
    {
        if (!$this->enoughMoney($money)) {
            return;
        }
        $this->balance->subtract($money);
    }
}

是否违反LSP?是的。这是因为帐户合同告诉我们帐户将被提取，但情况并非总是如此。那么，我该怎么做才能解决这个问题呢?我只是修改了合同:

interface Account
{
    /**
     * Withdraw $money amount from this account if its balance is enough.
     * Otherwise do nothing.
     *
     * @param Money $money
     * @return mixed
     */
    public function withdraw(Money $money);
}

Voilà，现在合同已得到满足。

这种微妙的违反通常会使客户有能力区分所使用的具体对象之间的差异。例如，给定第一个Account的契约，它看起来像下面这样:

class Client
{
    public function go(Account $account, Money $money)
    {
        if ($account instanceof DefaultAccount && !$account->hasEnoughMoney($money)) {
            return;
        }
        $account->withdraw($money);
    }
}

而且，这自动违反了开闭原则(即取款要求)。因为你永远不知道如果违反合同的对象没有足够的钱会发生什么。它可能什么都不返回，可能会抛出异常。所以你必须检查它是否hasEnoughMoney()——这不是接口的一部分。因此这种强制的依赖于具体类的检查违反了OCP。

这一点也解决了我经常遇到的关于LSP违反的误解。它说:“如果父母的行为在孩子身上改变了，那么它就违反了LSP。”然而，事实并非如此——只要孩子不违反父母的契约。

LSP是关于类的契约的规则:如果基类满足契约，则LSP派生的类也必须满足该契约。

在Pseudo-python

class Base:
   def Foo(self, arg): 
       # *... do stuff*

class Derived(Base):
   def Foo(self, arg):
       # *... do stuff*

如果每次在派生对象上调用Foo，它给出的结果与在Base对象上调用Foo完全相同，只要arg是相同的。

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. [...]

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。

当一些代码认为它正在调用类型T的方法时，LSP是必要的，并且可能在不知情的情况下调用类型S的方法，其中S扩展了T(即S继承、派生于超类型T，或者是超类型T的子类型)。

例如，当一个函数的输入形参类型为T时，调用(即调用)的实参值类型为S。或者，当一个类型为T的标识符被赋值类型为S时，就会发生这种情况。

val id : T = new S() // id thinks it's a T, but is a S

LSP要求T类型方法(例如Rectangle)的期望(即不变量)，当调用S类型方法(例如Square)时不违反此期望。

val rect : Rectangle = new Square(5) // thinks it's a Rectangle, but is a Square
val rect2 : Rectangle = rect.setWidth(10) // height is 10, LSP violation

即使是具有不可变字段的类型仍然有不变量，例如，不可变的矩形设置器期望维度被独立修改，但不可变的正方形设置器违背了这一期望。

class Rectangle( val width : Int, val height : Int )
{
   def setWidth( w : Int ) = new Rectangle(w, height)
   def setHeight( h : Int ) = new Rectangle(width, h)
}

class Square( val side : Int ) extends Rectangle(side, side)
{
   override def setWidth( s : Int ) = new Square(s)
   override def setHeight( s : Int ) = new Square(s)
}

LSP要求子类型S的每个方法必须有逆变的输入参数和协变的输出。

逆变是指方差与继承方向相反，即子类型S的每个方法的每个输入参数的Si类型必须与超类型T的相应方法的相应输入参数的Ti类型相同或为超类型。

协方差是指子类型S的每个方法的输出的方差在继承的同一方向，即类型So，必须是超类型T的相应方法的相应输出的相同或类型To的子类型。

这是因为如果调用者认为它有一个类型T，认为它正在调用一个类型T的方法，那么它就会提供类型Ti的参数，并将输出分配给类型to。当它实际调用S的对应方法时，每个Ti输入参数被赋值给Si输入参数，So输出被赋值给类型to。因此，如果Si与Ti的w.r.t.不是逆变的，那么就可以将Si的子类型xi赋给Ti，而它不是Si的子类型。

此外，对于在类型多态性参数(即泛型)上具有定义-站点方差注释的语言(例如Scala或Ceylon)，类型T的每个类型参数的方差注释的共方向或反方向必须分别与具有类型参数类型的每个输入参数或输出(T的每个方法)的方向相反或相同。

此外，对于每个具有函数类型的输入参数或输出，所需的方差方向是相反的。该规则是递归应用的。

子类型适用于可以枚举不变量的地方。

关于如何对不变量建模，以便由编译器强制执行，有很多正在进行的研究。

Typestate (see page 3) declares and enforces state invariants orthogonal to type. Alternatively, invariants can be enforced by converting assertions to types. For example, to assert that a file is open before closing it, then File.open() could return an OpenFile type, which contains a close() method that is not available in File. A tic-tac-toe API can be another example of employing typing to enforce invariants at compile-time. The type system may even be Turing-complete, e.g. Scala. Dependently-typed languages and theorem provers formalize the models of higher-order typing.

Because of the need for semantics to abstract over extension, I expect that employing typing to model invariants, i.e. unified higher-order denotational semantics, is superior to the Typestate. ‘Extension’ means the unbounded, permuted composition of uncoordinated, modular development. Because it seems to me to be the antithesis of unification and thus degrees-of-freedom, to have two mutually-dependent models (e.g. types and Typestate) for expressing the shared semantics, which can't be unified with each other for extensible composition. For example, Expression Problem-like extension was unified in the subtyping, function overloading, and parametric typing domains.

我的理论立场是，对于知识的存在(见章节“集中化是盲目的和不合适的”)，永远不会有一个通用模型可以在图灵完备的计算机语言中强制100%覆盖所有可能的不变量。要让知识存在，就必须存在许多意想不到的可能性，即无序和熵必须总是在增加。这是熵力。证明一个潜在扩展的所有可能的计算，就是计算一个先验的所有可能的扩展。

This is why the Halting Theorem exists, i.e. it is undecidable whether every possible program in a Turing-complete programming language terminates. It can be proven that some specific program terminates (one which all possibilities have been defined and computed). But it is impossible to prove that all possible extension of that program terminates, unless the possibilities for extension of that program is not Turing complete (e.g. via dependent-typing). Since the fundamental requirement for Turing-completeness is unbounded recursion, it is intuitive to understand how Gödel's incompleteness theorems and Russell's paradox apply to extension.

对这些定理的解释将它们纳入对熵力的广义概念理解中:

Gödel's incompleteness theorems: any formal theory, in which all arithmetic truths can be proved, is inconsistent. Russell's paradox: every membership rule for a set that can contain a set, either enumerates the specific type of each member or contains itself. Thus sets either cannot be extended or they are unbounded recursion. For example, the set of everything that is not a teapot, includes itself, which includes itself, which includes itself, etc…. Thus a rule is inconsistent if it (may contain a set and) does not enumerate the specific types (i.e. allows all unspecified types) and does not allow unbounded extension. This is the set of sets that are not members of themselves. This inability to be both consistent and completely enumerated over all possible extension, is Gödel's incompleteness theorems. Liskov Substition Principle: generally it is an undecidable problem whether any set is the subset of another, i.e. inheritance is generally undecidable. Linsky Referencing: it is undecidable what the computation of something is, when it is described or perceived, i.e. perception (reality) has no absolute point of reference. Coase's theorem: there is no external reference point, thus any barrier to unbounded external possibilities will fail. Second law of thermodynamics: the entire universe (a closed system, i.e. everything) trends to maximum disorder, i.e. maximum independent possibilities.

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?”