Question

我正在使用TypeScript中的专用函数签名。我的理解是以下应该有效：

// Basic superclass and subclass
class Model { }
class Product extends Model { name: string; }

// Define interface for singleton factory class with generic create method.
interface IModels { create(type: any): Model; }

const ctors = {product: Product};

// Implement generic version of create.
class Models implements IModels {
  create(type: any): Model { return new ctors[type](); }
}

// Extend interface with specialized signature.
interface IModels { create(type: "product"): Product; }

const models = new Models;

const product: Product = models.create("product");

但是，这会在最后一行产生以下错误：

Class 'Models' incorrectly implements interface 'IModels'.
  Types of property 'create' are incompatible.
    Type '(type: any) => Model' is not assignable to type '{ (type: any): Model; (type: "product"): Product; }'.
      Type 'Model' is not assignable to type 'Product'.
        Property 'name' is missing in type 'Model'.

如果我将create的返回类型从Model更改为any，那么它会编译，但为什么我必须这样做？

Answer 1

界面......

interface IModels { create(type: "product"): Product; }

...通过方法签名... {/ p>在您的Models类中实现

create(type: any): Model;

...因为any是"product"的可接受实现。然后，编译器仅使用Models类中的方法信息。

您还可以通过将此信息添加到Models类以及过载签名中来消除错误：

class Models implements IModels {
    create(type: "product"): Product;
    create(type: any): Model;
    create(type: any): Model {
        return new ctors[type]();
    }
}

TypeScript

1 个答案: