泛型 class 的 Curried 构造函数

Question

首先让我说我不确定这是否可能。

我正在尝试获取一个可以用 new 调用的构造函数，它不带参数，调用通用 class 的带参数的构造函数。像这样：

class SimpleFoo {
    public Key: String = null;
}

class GenericFoo<T> {
    public Key: T = null;
    constructor(private type: { new (): T }) {
        this.Key = new this.type();
    }
}

let simpleCtor: { new (): SimpleFoo } = SimpleFoo; // works
let simpleObj = new simpleCtor();

let genericCtor: { new (): GenericFoo<String> } = GenericFoo<String>(String); // <-- non-working code -- how to curry String parameter?
let genericObj = new genericCtor(); // this is how I wish to get a new object, no parameters involved

Answer 1

我不确定你想做什么，但这似乎有效：

type TypeConstructor<T> = { new (): T };

class GenericFoo<T> {
    public Key: T = null;
    constructor(private type: TypeConstructor<T>) {
        this.Key = new this.type();
    }
}

let genericCtor: { new (type: TypeConstructor<String>): GenericFoo<String> } = GenericFoo;
let genericObj = new genericCtor(String);

(code in playground)

编辑

如果你不想在调用构造函数时传递类型，那么你可以将构造函数绑定到想要的类型，然后调用绑定的构造函数：

type BoundGenericFooConstructor<T> = { new(): GenericFoo<T> }

let genericCtor: BoundGenericFooConstructor<String> = GenericFoo.bind(null, String);
let genericObj = new genericCtor();

(code in playground)

泛型 class 的 Curried 构造函数

Curried constructor of generic class

constructor

currying

typescript

编辑