到目前为止，我发现LSP最清晰的解释是“利斯科夫替换原则说，派生类的对象应该能够替换基类的对象，而不会给系统带来任何错误，也不会修改基类的行为”。文中给出了违反LSP的代码示例并进行了修复。

利斯科夫替换原理

(固体)

继承子类型化

维基里斯科夫替换原理(LSP)

在子类型中不能加强先决条件。 后置条件不能在子类型中减弱。 超类型的不变量必须保留在子类型中。

子类型不应该要求调用者提供比超类型更多的(先决条件) 子类型不应该为小于超类型的调用者公开(后置条件)

*前置条件+后置条件=函数(方法)类型[Swift函数类型。Swift函数与方法

//Swift function
func foo(parameter: Class1) -> Class2

//function type
(Class1) -> Class2

//Precondition
Class1

//Postcondition
Class2

例子

//C3 -> C2 -> C1

class C1 {}
class C2: C1 {}
class C3: C2 {}

前提条件(如。函数参数类型)可以相同或更弱(力求-> C1) 后置条件(如。函数返回类型)可以相同或更强(力求-> C3) 超类型的不变变量[About]应该保持不变

斯威夫特

class A {
    func foo(a: C2) -> C2 {
        return C2()
    }
}

class B: A {
    override func foo(a: C1) -> C3 {
        return C3()
    }
}

Java

class A {
    public C2 foo(C2 a) {
        return new C2();
    }
}

class B extends A {
    @Override
    public C3 foo(C2 a) { //You are available pass only C2 as parameter
        return new C3();
    }
}

行为子类型化

维基里斯科夫替换原理(LSP)

子类型中方法参数类型的逆变性。子类型中方法返回类型的协方差。 子类型中的方法不能引发新的异常，除非它们是超类型的方法引发的异常的子类型。

[方差，协方差，逆变，不变性]

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.

设q(x)是关于类型为T的x的对象的可证明属性，那么q(y)对于类型为S的对象y应该是可证明的，其中S是T的子类型。

实际上，公认的答案并不是利斯科夫原理的反例。正方形自然是一个特定的矩形，因此从类矩形继承是完全有意义的。你只需要以这样的方式实现它:

@Override
public void setHeight(double height) {
   this.height = height;
   this.width = height; // since it's a square
}

@Override
public void setWidth(double width) {
   setHeight(width);
}

所以，提供了一个很好的例子，然而，这是一个反例:

class Family:
-- getChildrenCount()

class FamilyWithKids extends Family:
-- getChildrenCount() { return childrenCount; } // always > 0

class DeadFamilyWithKids extends FamilyWithKids:
-- getChildrenCount() { return 0; }
-- getChildrenCountWhenAlive() { return childrenCountWhenAlive; }

在这个实现中，DeadFamilyWithKids不能从FamilyWithKids继承，因为getChildrenCount()返回0，而从FamilyWithKids它应该总是返回大于0的值。

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。

假设我们在代码中使用了一个矩形

r = new Rectangle();
// ...
r.setDimensions(1,2);
r.fill(colors.red());
canvas.draw(r);

在几何课上，我们学过正方形是一种特殊类型的矩形，因为它的长宽相等。让我们根据下面的信息创建一个Square类:

class Square extends Rectangle {
    setDimensions(width, height){
        assert(width == height);
        super.setDimensions(width, height);
    }
} 

如果我们在第一个代码中将矩形替换为正方形，那么它将会中断:

r = new Square();
// ...
r.setDimensions(1,2); // assertion width == height failed
r.fill(colors.red());
canvas.draw(r);

这是因为正方形有一个我们在矩形类中没有的新前提条件:width == height。根据LSP，矩形实例应该被矩形子类实例替代。这是因为这些实例通过了矩形实例的类型检查，因此它们将在代码中导致意外错误。

这是wiki文章中“在子类型中不能加强先决条件”部分的一个例子。因此，总而言之，违反LSP可能会在某些时候导致代码错误。