流未检测块是否处理潜在的未定义(可能)值。为什么?

我看到了一种我不理解的行为。我要键入的函数是redux reducer。我看到的问题似乎源于 action 提供给减速机的对象具有标记为“可能”运算符的有效负载: ?

type Action = {
  type: 'foo',
  payload?: { ... }
}

在减速器中,我试图提供一个if门来处理 payload 未定义。从理论上讲,这似乎是可行的。最简单的情况是可行的(参见 here ):

type Foo = {
  type: 'foo',
  payload?: {
    foo: 'foo'
  }
};

type Bar = {
  type: 'bar',
  payload?: {
    bar: 'bar',
  }
};

const reducer = (state: {} = {}, action: Foo | Bar) => {
  switch (action.type) {
    case 'foo': {
      if (!action.payload) {
        return state;
      }
      return action.payload.foo;
    }
    case 'bar': {
      if (!action.payload) {
        return state;
      }
      return action.payload.bar;
    }
    default:
      return state;
  }
}

然而,在我的实际减速机,这是一个有点复杂,我无法摆脱的错误。我的减速机看起来像这样:

const camelize = (x) => x;
const getSearchTypeFromPath = (x) => x;

type A = {
  type: '@@router/LOCATION_CHANGE',
  payload?: {
    pathname: string,
    query: {},
  }
}

type B = {
  type: 'action.foo',
  payload?: {
    mode: string,
    params: {}
  }
}

const byMode = (
  state: {} = {},
  action: A | B,
) => {
  switch (action.type) {
    case '@@router/LOCATION_CHANGE': {
      if (!action.payload) {
        return state;
      }

      const modeKey: string = camelize(getSearchTypeFromPath(action.payload.pathname));
      return {
        ...state,
        [modeKey]: action.payload.query,
      };
    }
    case 'action.foo': {
      if (!action.payload) {
        return state;
      }

      const modeKey: string = camelize(action.payload.mode);
      return {
        ...state,
        [modeKey]: action.payload.params,
      }
    }
    default:
      return state;
  }
};

与上面的简化情况不同,此代码会产生一些流错误:请参见 here :

33:         [modeKey]: action.payload.query,
                                      ^ Cannot get `action.payload.query` because property `query` is missing in undefined [1].
References:
6:   payload?: {               ^ [1]
44:         [modeKey]: action.payload.params,
                                      ^ Cannot get `action.payload.params` because property `params` is missing in undefined [1].
References:
14:   payload?: {                ^ [1]

有效载荷 我也这么认为。为什么会出错?有趣的是,每个case块中引用的第一个属性没有错误:即。 pathname 和 mode . 最后,我注意到的另一件事是,如果我删除helper函数,在我的示例中,这些函数刚刚变成了伪函数,那么我就不会得到错误。见 here .