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.

简单来说，LSP是指同一超类的对象应该能够在不破坏任何东西的情况下相互交换。

例如，如果我们有一个从Animal类派生的Cat和Dog类，那么任何使用Animal类的函数都应该能够使用Cat或Dog，并且行为正常。

在一个非常简单的句子中，我们可以说:

子类不能违背它的基类特征。它必须有能力。我们可以说这和子类型是一样的。

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。

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。

利斯科夫替换原理

被重写的方法不应该保持为空 被重写的方法不应该抛出错误 基类或接口行为不应该因为派生类行为而进行修改(重做)。