Typescript 限制返回联合的TS函数的返回类型

我编写了一个util函数，用于解析API异常并返回建模的TS类型.

interface NotFoundException {
  type: 'notFound'
}

interface PermissionException {
  type: 'permission'
}

interface BadRequestException {
  type: 'badRequest'
}

// Example shape of API Error 
interface APIError {
  statusCode: number;
}

type KnownExceptions = NotFoundException | PermissionException | BadRequestException;

function exceptionHandler(err: unknown): KnownExceptions {
  const statusCode = (err as APIError).statusCode ?? -1;

  if (statusCode === 404) {
    return {
      type: 'notFound'
    }
  }

  if (statusCode === 403) {
    return {
      type: 'permission'
    }
  }

  if (statusCode === 400) {
    return {
      type: 'badRequest'
    }
  }

  throw new Error('Unknown error', { cause: err });
}

有些API不会返回所有类型，因此我需要显式处理其他类型

interface APIInput {
  prop: string;
}

interface APIOutput {
  id: string;
}

function callAPI(input: APIInput): APIOutput | PermissionException | NotFoundException {
  try {
    //Mock API call
    return {
      id: 'id'
    };
  } catch (err: unknown) {
    const e = exceptionHandler(err);

    // Problem: This API (and others) cannot throw this/some exceptions

    // Need a function that will restrict the types returned by exceptionHandler
    // and throw 'Unexpected exception' for impossible cases
    if (e.type === 'badRequest') {
      throw new Error('Unexpected exception', { cause: err });
    }

    return e;
  }
}

Playground Link

有没有一种方法可以让我写一个函数，将异常联合的返回类型限制为我想要的类型，并对所有其他类型抛出Unexpected exception个错误？

由于TS不能将类型作为参数，我发现很难弄清楚这一点.