最近尝试用TypeScript写一个工具库，需要实现这样一个场景：
声明一个抽象类Parent
声明一组子类ChildA、ChildB继承这个Parent，实现它的抽象方法
实现一个方法，根据参数返回对应的子类
用拿到的子类创建实例
代码示例如下：
typescript
abstract class Animal {
abstract makeSound(): void
}
class Dog extends Animal {
makeSound(): void {
console.log('woof')
}
}
class Cat extends Animal {
makeSound(): void {
console.log('meow')
}
}
const getAnimal = (name: string) => {
if (name === 'cat') return Cat
return Dog
}
const animal = new (getAnimal('dog'))()
animal.makeSound() // woofabstract class Animal {
abstract makeSound(): void
}
class Dog extends Animal {
makeSound(): void {
console.log('woof')
}
}
class Cat extends Animal {
makeSound(): void {
console.log('meow')
}
}
const getAnimal = (name: string) => {
if (name === 'cat') return Cat
return Dog
}
const animal = new (getAnimal('dog'))()
animal.makeSound() // woof首先注意new后面getAnimal方法的执行需要用括号包起来，否则将得到以下错误：
TS2350: Only a void function can be called with the 'new' keyword.
ESLint: A constructor name should not start with a lowercase letter.TS2350: Only a void function can be called with the 'new' keyword.
ESLint: A constructor name should not start with a lowercase letter.随后按照严谨的做法，我尝试给这个getAnimal方法添加类型约束：
typescript
const getAnimal = (name: string): Animal => {
if (name === 'cat') return Cat
return Dog
}const getAnimal = (name: string): Animal => {
if (name === 'cat') return Cat
return Dog
}马上得到了错误提示：
TS2351: This expression is not constructable.
Type 'Animal' has no construct signatures.TS2351: This expression is not constructable.
Type 'Animal' has no construct signatures.这样写的错误在于，Animal描述的应当是一个由其创建的实例的类型（或者说类）。
比如改写成下面这样就没有问题了:
typescript
const getAnimalInstance = (name: string): Animal => {
if (name === 'cat') return new Cat()
return new Dog()
}const getAnimalInstance = (name: string): Animal => {
if (name === 'cat') return new Cat()
return new Dog()
}而上面的getAnimal方法返回的不是实例，是类（构造器）本身。
描述这种类型，需要用到TypeScript的new ()语法
typescript
const getAnimal = (name: string): { new (): Animal } => {
if (name === 'cat') return Cat
return Dog
}const getAnimal = (name: string): { new (): Animal } => {
if (name === 'cat') return Cat
return Dog
}或者这样写：
typescript
const getAnimal = (name: string): new () => Animal => {
if (name === 'cat') return Cat
return Dog
}const getAnimal = (name: string): new () => Animal => {
if (name === 'cat') return Cat
return Dog
}表示这个方法返回的是一个构造器，这个构造器可以创造出一个类型是Animal的实例。
最后的示例如下：
typescript
abstract class Animal {
abstract makeSound(): void
}
class Dog extends Animal {
makeSound(): void {
console.log('woof')
}
}
class Cat extends Animal {
makeSound(): void {
console.log('meow')
}
}
const getAnimal = (name: string): { new (): Animal } => {
if (name === 'cat') return Cat
return Dog
}
const getAnimalInstance = (name: string): Animal => {
if (name === 'cat') return new Cat()
return new Dog()
}
console.log('sound:', new (getAnimal('dog'))().makeSound())
console.log('sound:', getAnimalInstance('cat').makeSound())abstract class Animal {
abstract makeSound(): void
}
class Dog extends Animal {
makeSound(): void {
console.log('woof')
}
}
class Cat extends Animal {
makeSound(): void {
console.log('meow')
}
}
const getAnimal = (name: string): { new (): Animal } => {
if (name === 'cat') return Cat
return Dog
}
const getAnimalInstance = (name: string): Animal => {
if (name === 'cat') return new Cat()
return new Dog()
}
console.log('sound:', new (getAnimal('dog'))().makeSound())
console.log('sound:', getAnimalInstance('cat').makeSound())打开在线示例查看执行结果：
https://codepen.io/mirari/pen/xxbrvVd
点击Console打开控制台
参考文档：