当使用扩展时,typescript是否可以像使用联合时一样推断开关块中的正确类型?

typescript推断 serverMessage 正确地作为 ServerInformation 在以下情况下:

export interface IServerResponse {
    type: 'response';
    cmd: string;
}

export interface ServerError {
    type: 'error',
    errorcode: string;
    cmd?: string;
    trans_id?: string;
    data?: Object;
}

export interface ServerInformation {
    type: 'information',
    info: string;
    data?: Object;
}

export type ServerMessage = IServerResponse | ServerError | ServerInformation;

let serverMessage = message as ServerMessage;

switch (serverMessage.type) {
    case 'information':
        const information: Information = serverMessage.info; // type is inferred correctly
        break;

但在我看来,它更符合我在其他地方用来扩展不同类型消息的类语法。 IServerMessage 比把它们组合成 ServerMessage 类型:

能做到吗?我在下面的尝试失败。

export interface IServerMessage {
    type: string;
}

export interface IServerResponse extends IServerMessage {
    type: 'response';
    cmd: string;
}

export interface ServerError extends IServerMessage {
    type: 'error',
    errorcode: string;
    cmd?: string;
    trans_id?: string;
    data?: Object;
}

export interface ServerInformation extends IServerMessage {
    type: 'information',
    info: string;
    data?: Object;
}

let serverMessage: IServerMessage = message ;

switch (serverMessage.type) {
    case 'information':
        const information: Information = serverMessage.info; // type is still inferred as ServerMessage, not the more specific InformationMessage
        break;