Typescript：多态的 this，重载签名与函数实现不兼容

有人能解释一下为什么这不起作用吗：

abstract class Model {
    static deserialize<T extends Model>(this: ( new() => T ), object: any): T;
    static deserialize<T extends Model>(ctor: ( new() => T ), object: any): T {
        return new ctor();
    }

    static deserializeArray<T extends Model>(this: ( new() => T ), ...objects: any[]): T[];
    static deserializeArray<T extends Model>(ctor: ( new() => T ), ...objects: any[]): T[] {
        return objects.map(object => Model.deserialize(ctor, object));
    }
}

class MyModel extends Model { }

这将允许：

let myModel = MyModel.deserialize({});
let myModels = MyModel.deserializeArray({}, {}, {});

或

let myModel = Model.deserialize(MyModel, {});
let myModels = Model.deserializeArray(MyModel, {}, {}, {});

Typescript 2.5.2 报告“重载签名与函数实现不兼容”。

为什么两种形式都需要？

考虑一个返回序列化（json）模型的 REST API：

class MyModelController {
    get(id: number) {
        let myModel = ... some db/service call ...
        return myModel.serialize();
    }
}

然后是一个通用的服务（Angular）用于请求模型：

@Injectable()
abstract class HttpService {
    constructor(private http: Http) { }

    errorHandler(response) {
        ...
    }

    get<T extends Model>(ModelType: ( new() => T ), endpoint: string): Observable<T> {
        return this.http.get(endpoint)
            // we can't call ModelType.deserialize() here...
            .map(response => Model.deserialize(ModelType, response.json()))
            .catch(response => this.errorHandler(response));
    }
}

@Injectable()
class MyModelService extends HttpService {
    get(id: number) {
       return super.get(MyModel, `/api/models/${id}`);
    }
}

解决方案

abstract class Model {
    static deserialize<T extends Model>(this: ( new() => T), object: {}): T;
    static deserialize<T extends Model>(this: Function & { prototype: Model }, ctor: ( new() => T ), object: {});
    static deserialize<T extends Model>(this: ( new() => T), first: ( new() => T ) | {}, second?: any) {
        return typeof first === "function" ? new first() : new this(); 
    }

    static deserializeArray<T extends Model>(this: ( new() => T), ...objects: {}[]): T[];
    static deserializeArray<T extends Model>(this: Function & { prototype: Model }, ctor: ( new() => T ), ...objects: {}[]): T[];
    static deserializeArray<T extends Model>(this: ( new() => T), first: ( new() => T ) | {}[], second?: {}[]): T[] {
        const ctor = typeof first === "function" ? first : this;
        const objects = typeof first === "function" ? second : first;
        return objects.map(object => Model.deserialize(ctor, object));
    }
}

这允许两种形式，同时保留 abstract。

答案＃1

在我看来，没有必要两者都要：

let myModel1 = MyModel.deserialize({});
let myModel2 = Model.deserialize(MyModel, {});

第一种形式就足够了，而且更易读，然后代码将如下所示：

type ModelCtor<T extends Model> = {
    new(): T;
    deserialize<T extends Model>(object: any): T;
    deserializeArray<T extends Model>(...objects: any[]): T[];
};

abstract class Model {
    static deserialize<T extends Model>(this: ModelCtor<T>, object: any): T {
        return new this();
    }

    static deserializeArray<T extends Model>(this: ModelCtor<T>, ...objects: any[]): T[] {
        return objects.map(object => this.deserialize(object));
    }
}

class MyModel extends Model { }

let myModel = MyModel.deserialize({});
let myModels = MyModel.deserializeArray({}, {}, {});

（在游乐场中的代码）

然而，如果出于某种我无法理解的原因，你确实想同时拥有这两种形式，那么你可以这样做：

type ModelCtor<T extends Model> = {
    new(): T;
    deserialize<T extends Model>(object: any): T;
    deserialize<T extends Model>(ctor: ModelCtor<T>, object: any): T;

    deserializeArray<T extends Model>(...objects: any[]): T[];
    deserializeArray<T extends Model>(ctor: ModelCtor<T>, ...objects: any[]): T[];
};

class Model {
    static deserialize<T extends Model>(this: ModelCtor<T>, object: any): T;
    static deserialize<T extends Model, S extends Model>(this: ModelCtor<T>, ctor: ModelCtor<S>, object: any): S;
    static deserialize<T extends Model, S extends Model>(this: ModelCtor<T>, first: ModelCtor<S> | any, second?: any): S {
        return second === undefined ? new this() : new first();
    }

    static deserializeArray<T extends Model>(this: ModelCtor<T>, ...objects: any[]): T[];
    static deserializeArray<T extends Model, S extends Model>(this: ModelCtor<T>, ctor: ModelCtor<S>, ...objects: any[]): S[];
    static deserializeArray<T extends Model, S extends Model>(this: ModelCtor<T>, ...objects: any[]): S[] {
        const ctor: ModelCtor<S> = typeof objects[0] === "function" ? objects[0] : this as any;
        return objects.map(object => ctor.deserialize(ctor, object));
    }
}

class MyModel extends Model { }

let myModel1 = MyModel.deserialize({});
let myModels1 = MyModel.deserializeArray({}, {}, {});

let myModel2 = Model.deserialize(MyModel, {});
let myModels2 = Model.deserializeArray(MyModel, {}, {}, {});

（在游乐场中的代码）

代码更复杂了（我认为没有充分的理由）。
另外，我不得不从 Model 中移除 abstract 部分，否则会出现这个错误：

类型为 'typeof Model' 的 'this' 上下文不能分配给类型为 'ModelCtor' 的方法的 'this'。
不能将抽象构造函数类型分配给非抽象构造函数类型。

对于这两行：

let myModel2 = Model.deserialize(MyModel, {});
let myModels2 = Model.deserializeArray(MyModel, {}, {}, {});

也许你可以再多试一试，并修复那些错误，而不删除abstract。