小编xva*_*aro的帖子

我试图使用foldr实现filterM函数,但是接收到一个我无法理解的错误.

我的实现(这只是为了理解,我知道我应该使用do notation ...):

filterM :: (Monad m) => (a -> (m Bool)) -> [a] -> m [a]
filterM f list = foldr foldFn (return []) list
    where 
        foldFn :: (Monad m) => a -> m [a] -> m [a]
        foldFn x acc = let 
            m = f x
            in acc >>= 
                \l -> m >>= 
                    \b -> (if b == True then return (x:l) else return l)

我收到以下错误

Could not deduce (a ~ a1)
from the context (Monad m) …

我在网上找到的用于缩小flowtype中不相交联合的所有示例都使用字符串文字,就像官方文字一样.我想知道是否有办法检查枚举中的值,如:

const ACTION_A = 'LITERAL_STRING_A';
const ACTION_B = 'LITERAL_STRING_B';

type ActionA = {
  // This is not allowed
  type: ACTION_A,
  // type: 'LITERAL_STRING_A' is allowed
  dataA: ActionAData,
}

type ActionB = {
  // This is not allowed
  type: ACTION_B,
  // type: 'LITERAL_STRING_B' is allowed
  dataB: ActionBData,
}

type Action = ActionA | ActionB;

function reducer(state: State, action: Action): State {
  // Want to narrow Action to ActionA or ActionB based on type
  switch (action.type) {
    // case 'LITERAL_STRING_A': …