TypeScript 通用工厂函数基于实例参数返回类型而不重载

Question

我想根据提供给工厂方法的实例参数创建实例。我能够使用重载编写以下代码。 如何使用 TypeScript 泛型来表达？

class Car {
    brand = 'Skoda';
}

class Human {
    name = 'Jan';
}

class Garage {
    lots = 5;
}

class House {
    floors = 2;
}

class NotAllowed {
    anything = 'anything';
}

class Selection {

    pick(what: Car) : Garage;
    pick(what: Human) : House;
    pick(what: any) : any {
        if (what instanceof Car) {
            return new Garage();
        } else if (what instanceof Human) {
            return new House();
        } else {
            throw Error('Not supported')
        }
    }

}

const selection = new Selection();
selection.pick(new Car()).lots; // compiler successfully autocompletes 
selection.pick(new Human()).floors // compiler successfully autocompletes 
selection.pick(new NotAllowed()); // compiler error

Answer 1

您可以将pick() 的调用签名更改为generic 方法，其参数为泛型类型constrained 为您所需的不同输入类型的union，其返回类型为@987654324 @根据输入类型选择输出类型：

declare class Selection {
  pick<T extends Car | Human>(what: T): T extends Car ? Garage : House;
}

这对于您的示例调用的行为相同：

const selection = new Selection();
selection.pick(new Car()).lots; // compiler successfully autocompletes 
selection.pick(new Human()).floors // compiler successfully autocompletes 
selection.pick(new NotAllowed()); // compiler error

---

很遗憾，您的实现不会被视为类型安全的； T 类型将在pick() 的主体内未指定，因此编译器不会也无法验证T 是否可分配给Car，即使what instanceof Car 为真：

pick<T extends Car | Human>(what: T): T extends Car ? Garage : House {
  if (what instanceof Car) {
    return new Garage(); // error!
    // Type 'Garage' is not assignable to type 'T extends Car ? Garage : House'
  } else if (what instanceof Human) {
    return new House(); // error!
    // Type 'House' is not assignable to type 'T extends Car ? Garage : House'.
  } else {
    throw Error('Not supported')
  }
}

我不会继续解释为什么会发生这种情况，我只会将您引导至 microsoft/TypeScript#33912，这是一个相关的开放功能请求，旨在改进语言支持以实现返回条件类型的泛型函数。

现在你需要做类似type assertion:

pick<T extends Car | Human>(what: T): T extends Car ? Garage : House {
  if (what instanceof Car) {
    return new Garage() as T extends Car ? Garage : House;
  } else if (what instanceof Human) {
    return new House() as T extends Car ? Garage : House;
  } else {
    throw Error('Not supported')
  }
}

或等效地，单个呼叫签名overload：

pick<T extends Car | Human>(what: T): T extends Car ? Garage : House;
pick(what: Car | Human) {
  if (what instanceof Car) {
    return new Garage();
  } else if (what instanceof Human) {
    return new House();
  } else {
    throw Error('Not supported')
  }
}

我知道你说过你不想重载，但这通常是我的解决方案，因为到处添加类型断言更麻烦也更乏味。

Playground link to code