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class Greeter {
greeting: string;
constructor(message: string) {
this.greeting = message;
}
greet() {
return "Hello, " + this.greeting;
}
}
let greeter = new Greeter("world");
class Animal {
name: string;
constructor(theName: string) { this.name = theName; }
move(distanceInMeters: number = 0) {
console.log(`${this.name} moved ${distanceInMeters}m.`);
}
}
class Snake extends Animal {
constructor(name: string) { super(name); }
move(distanceInMeters = 5) {
console.log("Slithering...");
super.move(distanceInMeters);
}
}
class Horse extends Animal {
constructor(name: string) { super(name); }
move(distanceInMeters = 45) {
console.log("Galloping...");
super.move(distanceInMeters);
}
}
let sam = new Snake("Sammy the Python");
let tom: Animal = new Horse("Tommy the Palomino");
sam.move();
tom.move(34);
如同C++的类型继承,
在TypeScript里,成员都默认为public
类型兼容
TypeScript使用的是结构性类型系统。 当我们比较两种不同的类型时,并不在乎它们从何处而来,如果所有成员的类型都是兼容的,我们就认为它们的类型是兼容的。
注意:类型兼容检查时,会检查类型名,类型,和声明处是否一致,均一致时才通过
你可以使用readonly关键字将属性设置为只读的。 只读属性必须在声明时或构造函数里被初始化。
class Octopus {
readonly name: string;
readonly numberOfLegs: number = 8;
constructor (theName: string) {
this.name = theName;
}
}
let dad = new Octopus("Man with the 8 strong legs");
dad.name = "Man with the 3-piece suit"; // 错误! name 是只读的.
class Animal {
constructor(private name: string) { }
move(distanceInMeters: number) {
console.log(`${this.name} moved ${distanceInMeters}m.`);
}
}
仅在构造函数里使用private name: string参数来创建和初始化name成员。 我们把声明和赋值合并至一处
let passcode = "secret passcode";
class Employee {
private _fullName: string;
get fullName(): string {
return this._fullName;
}
set fullName(newName: string) {
if (passcode && passcode == "secret passcode") {
this._fullName = newName;
}
else {
console.log("Error: Unauthorized update of employee!");
}
}
}
let employee = new Employee();
employee.fullName = "Bob Smith";
if (employee.fullName) {
alert(employee.fullName);
}
注意:
- 存取器要求你将编译器设置为输出ECMAScript 5或更高
- 只带有get不带有set的存取器自动被推断为readonly
class Grid {
static origin = {x: 0, y: 0};
calculateDistanceFromOrigin(point: {x: number; y: number;}) {
let xDist = (point.x - Grid.origin.x);
let yDist = (point.y - Grid.origin.y);
return Math.sqrt(xDist * xDist + yDist * yDist) / this.scale;
}
constructor (public scale: number) { }
}
let grid1 = new Grid(1.0); // 1x scale
let grid2 = new Grid(5.0); // 5x scale
console.log(grid1.calculateDistanceFromOrigin({x: 10, y: 10}));
console.log(grid2.calculateDistanceFromOrigin({x: 10, y: 10}));
类声明为抽象类,定义抽象方法
abstract class Animal { abstract makeSound(): void; move(): void { console.log('roaming the earch...'); } }
抽象类中的抽象方法不包含具体实现并且必须在派生类中实现。 抽象方法的语法与接口方法相似。 两者都是定义方法签名但不包含方法体。 然而,抽象方法必须包含abstract关键字并且可以包含访问修饰符
abstract class Department {
constructor(public name: string) {
}
printName(): void {
console.log('Department name: ' + this.name);
}
abstract printMeeting(): void; // 必须在派生类中实现
}
class AccountingDepartment extends Department {
constructor() {
super('Accounting and Auditing'); // 在派生类的构造函数中必须调用 super()
}
printMeeting(): void {
console.log('The Accounting Department meets each Monday at 10am.');
}
generateReports(): void {
console.log('Generating accounting reports...');
}
}
let department: Department; // 允许创建一个对抽象类型的引用
department = new Department(); // 错误: 不能创建一个抽象类的实例
department = new AccountingDepartment(); // 允许对一个抽象子类进行实例化和赋值
department.printName();
department.printMeeting();
department.generateReports(); // 错误: 方法在声明的抽象类中不存在
注意:上栗存在与C++多态表现一致的行为,即子类实例赋值给基类引用,将作为父类实例使用,也就是对象上转,对象下转一般不被允许,不知道类型断言可不可以
把类当做接口使用
如上一节里所讲的,类定义会创建两个东西:类的实例类型和一个构造函数。 因为类可以创建出类型,所以你能够在允许使用接口的地方使用类
class Point {
x: number;
y: number;
}
interface Point3d extends Point {
z: number;
}
let point3d: Point3d = {x: 1, y: 2, z: 3};