Haskell:预期的懒惰，为什么要评估？

Question

我有一个功能 sideH这冒着 Prelude.head [] 的风险.因此，我使用Maybe来编写它，以避免这种情况:

sideH :: Residue -> Maybe (Atom, Atom)
sideH res
    -- Make sure the elements exist
    | nits /= [] && cars /= [] && oxys /= [] = Just (newH1, newH2)
    | otherwise = Nothing where
    ...

以上完全按预期工作，没有错误。现在，在调用 sideH 的函数中(这不是 do 构造)，我必须处理 sideH 的情况返回 Nothing :

callerFunc :: [Residue] -> Aromatic -> [(Double, Double)]
callerFunc [] _ = []
callerFunc (r:rs) aro
    -- Evaluate only if there is something to evaluate
    | newHs /= Nothing = (newH1Pos, newH2Pos)
    | otherwise = callerFunc rs aro where
    newHs = sideH r
    newH1Pos = atomPos $ fst $ fromJust newHs
    newH2Pos = atomPos $ snd $ fromJust newHs

如果我尝试评估 newH1Pos或 newH2Pos当 newH = Nothing ，它会失败，因为 fromJust Nothing是一个错误。但是，我希望这永远不会发生。我期待 callerFunc评估 newHs ，即 Just something或 Nothing .如果是 Nothing ，然后是 callerFunc将进入下一步而不评估 newH1Pos或 newH2Pos .情况似乎并非如此。我得到一个 *** Exception: Maybe.fromJust: Nothing错误正是我所期望的 newHs返回 Nothing .

我被要求提供更多代码。我试图想出一个重现错误的最小情况，但与此同时，这里是完整的问题 callerFunc代码。

-- Given a list of residues and an aromatic, find instances where there
--  is a Hydrogen bond between the aromatic and the Hydrogens on Gln or Asn
callerFunc :: [Residue] -> Aromatic -> [(Double, Double)]
callerFunc [] _ = []
callerFunc (r:rs) aro
    -- GLN or ASN case
    | fst delR <= 7.0 && (resName r == gln || resName r == asn) &&
        newHs /= Nothing && snd delR <= 6.0 = 
        [(snd delR, fst delR)] ++ hBondSFinder rs aro
    | otherwise = hBondSFinder rs aro where
    -- Sidechain identifying strings
    gln = B.pack [71, 76, 78]
    asn = B.pack [65, 83, 78]
    -- Get the location of the Hydrogens on the residue's sidechain
    newHs = sideH r
    newH1Pos = atomPos $ fst $ fromJust newHs
    newH2Pos = atomPos $ snd $ fromJust newHs
    -- Get the location of the Nitrogen on the mainchain of the Residue
    ats = resAtoms r
    backboneNPos = atomPos $ head $ getAtomName ats "N"
    hNVect1 = Line2P {lp1 = newH1Pos, lp2 = backboneNPos}
    hNVect2 = Line2P {lp1 = newH2Pos, lp2 = backboneNPos}
    interPoint1 = linePlaneInter (aroPlane aro) hNVect1
    interPoint2 = linePlaneInter (aroPlane aro) hNVect2
    delR = minimum [(interPoint1 `dist` newH1Pos, delr1), 
        (interPoint2 `dist` newH2Pos, delr2)]
    delr1 = interPoint1 `dist` (aroCenter aro)
    delr2 = interPoint2 `dist` (aroCenter aro)

我知道这是一个痛苦的代码转储。我正在努力减少它。

Answer 1

这个问题的答案(在评论中提出)不适合评论:“我不确定在这里如何使用模式匹配来删除这些 if 语句。”。

像这样，例如，尽管仍然存在一些代码异味，但可以通过一些额外的重构来改进:

sideH :: Residue -> Maybe (Atom, Atom)
sideH res = case (nits, cars, oxys) of
    (_:_, _:_, _:_) -> Just (newH1, newH2)
    _ -> Nothing
    where
    ...

如果你有灵活的道德，你可以尝试这样的事情:

sideH :: Residue -> Maybe (Atom, Atom)
sideH res = do
    _:_ <- return nits
    _:_ <- return cars
    _:_ <- return oxys
    return (newH1, newH2)
    where
    ...

同样，如果有更多的上下文和代码可用于提出建议，那么这两个代码示例可能会改进大约十倍。