哈斯克爾：函數薪酬返回Maybe類型：異常與fromJust功能

Question

我在Haskell

payment :: [(Integer , Integer)] -> Integer -> [(Integer , Integer)] -> Maybe [(Integer , Integer)] 
payment portemonnaie amount cashier | content (fromJust (pay portemonnaie amount)) - amount == 0 = Just [] 
            | pay portemonnaie amount == Nothing = Nothing 
            | otherwise = payExact cashier (content (fromJust (pay portemonnaie amount)) - amount) 

這裏的問題是寫了下面的功能。通過詢問

*** Exception: Maybe.fromJust: Nothing 

我試圖阻止：所以當功能「買單」返回Nothing我得到下面的異常

| pay portemonnaie amount == Nothing = Nothing 

但顯然這是不可能的。有人有更聰明的方式來構建這個？謝謝

Answer 1

問題是，Nothing檢查後電話fromJust。在你解決它的方式，它應該是這樣的：

payment :: [(Integer , Integer)] -> Integer -> [(Integer , Integer)] -> Maybe [(Integer , Integer)] 
payment portemonnaie amount cashier | pay portemonnaie amount == Nothing = Nothing 
            | content (fromJust (pay portemonnaie amount)) - amount == 0 = Just [] 
            | otherwise = payExact cashier (content (fromJust (pay portemonnaie amount)) - amount) 

然而，它可能仍然會更好。首先，自2010年以來哈斯克爾（或在Haskell 98與PatternGuards擴展名），它更容易寫：

payment :: [(Integer , Integer)] -> Integer -> [(Integer , Integer)] -> Maybe [(Integer , Integer)] 
payment portemonnaie amount cashier | Nothing <- pay portemonnaie amount = Nothing 
            | Just paid <- pay portemonnaie amount, content paid - amount == 0 = Just [] 
            | otherwise = payExact cashier (content paid - amount) 

此外，您可以通過使用where條款縮短一些事情：

payment :: [(Integer , Integer)] -> Integer -> [(Integer , Integer)] -> Maybe [(Integer , Integer)] 
payment portemonnaie amount cashier | Nothing <- mayHavePaid = Nothing 
            | Just paid <- mayHavePaid, content paid - amount == 0 = Just [] 
            | otherwise = payExact cashier (content paid - amount) 
    where 
    mayHavePaid = pay portemonnaie amount 

Answer 2

如果沒有任何擴展，仍然可以用相當可讀的方式來編寫（實際上，我認爲，比使用模式守護的答案更易讀）。首先，注意到一件事，你在所有的情況下做的是pay portemonnaie amount，那你必須做取決於函數的結果不同的東西：所以，把它放在第一，和模式匹配就可以了：

case pay portemonnaie amount of 
    -- ... 

在沒有的情況下是很容易，所以你可以平凡添加進來，並留下一個空白的只是：現在

case pay portemonnaie amount of 
    Nothing -> Nothing 
    (Just receipt) -> -- ... 

，如果有一個剛（這我在這裏呼籲收據，因爲我不是確定它在你的域中是什麼），那麼你總是需要計算剩餘的數量，所以讓我們把它放在where子句中：

case pay portemonnaie amount of 
    Nothing -> Nothing 
    (Just receipt) -> -- ... 
    where outstanding = content receipt - amount 

現在，所有剩下的就是填補了不同的保護條款對於剛剛模式，這是現在很容易，相關數值已經綁定：

case pay portemonnaie amount of 
    Nothing -> Nothing 
    (Just receipt) | outstanding == 0 -> Just [] 
       | otherwise -> payExact cashier outstanding 
    where outstanding = content receipt - amount 

當然插槽，表達回實際功能的主體：

payment :: [(Integer, Integer)] -> Integer -> [(Integer, Integer)] -> Maybe [(Integer, Integer)] 
payment portemonnaie amount cashier = 
    case pay portemonnaie amount of 
    Nothing -> Nothing 
    (Just receipt) | outstanding == 0 -> Just [] 
        | otherwise -> payExact cashier outstanding 
     where outstanding = content receipt - amount