typescript - 为什么 typescript 枚举无助于推断联合类型

Question

我得到了以下类，这是简单的 pubsub 实现:

interface IAction<N, T> {
    name: N,
    data: T
}

export class Communicator<A extends IAction<any, any>> {

    public subscribe(
       subscriber: { name: A['name']; handler: (data: A['data']) => void; }
    ) {}

    public emit(name: A['name'], data: A['data']): void {
    }
}

然后我像这样使用它:

enum ActionName {
    name1 = 'name1',
    name2 = 'name2'
}

type ExampleActions
    = { name: ActionName.name1, data: string }
    | { name: ActionName.name2, data: number };

const communicator = new Communicator<ExampleActions>();

let a = '';

communicator.subscribe({
    name: ActionName.name1,
    /* Expected no error, but get 
       TS2345: Type 'number' is not assignable to type 'string'.*/
    handler: data => a = data
});

/* Expected error Type 'string' is not assignable to type 'number',
    but got no error*/
communicator.emit(ActionName.name2, 'string'); 

看起来 typescript 可能会通过具体的枚举推断出一种具体的 Action，但事实并非如此。这是一个错误，还是我应该做些不同的事情？

Answer 1

问题是 A 是联合类型，所以 A['data'] 将是 string | number(联合中的所有可能性)。也没有推断根据 A['name'] 的类型确定 data 的类型应该不同，其他两者之间没有关系它们是联合字段的所有可能类型。

如果你在 typescript 2.8 中使用条件类型(在撰写本文时尚未发布，但你可以在 npm install -g typescript@next 上获得它，你可以获得期望的结果，它将被发布三月)

// Helper type TActions will be a union of any number of actions 
// TName will be the name of the action we are intresed in 
// We use filter, to filter out of TActions  all actions that are not named TName 
type ActionData<TActions extends IAction<string, any>, TName> = 
           Filter<TActions, IAction<TName, any>>['data']; 

export class Communicator<A extends IAction<any, any>> {

    public subscribe<K extends A['name']>(
    subscriber: { name: K; handler: (data: ActionData<A, K>) => void; }
    ) {}

    public emit<K extends A['name']>(name: K, data: ActionData<A,K>): void {
    }
}
const communicator = new Communicator<ExampleActions>();

let a = '';

communicator.subscribe({ name: ActionName.name1, handler: d=> a = d}); // ok 
communicator.subscribe({ name: ActionName.name2, handler: d=> a = d}); // erorr as it should be

communicator.emit(ActionName.name1, a); // ok 
communicator.emit(ActionName.name2, a); // erorr as it should be