在这个例子中，为什么天花板和地板操作员之间没有区别？ （Haskell的） [英] Why isn't there a difference between the ceiling and floor operators in this example? (Haskell)

问题描述

我必须将列表分成两半，导致列表的前半部分和列表的后半部分（half1，half2）的元组。当这个列表的长度是奇数时，我想让 half1 包含比 half2 更多的元素。

I have to split a list in half, resulting in a tuple of the first half of the list and the second half of the list (half1, half2). When the length of this list is odd, then I want to make half1 contain more elements than half2.

 halve :: [a] -> ([a],[a])
 halve [] = ([],[])
 halve xs =
    if (((length(xs) `mod` 2)==1))
    then(take (ceiling(toRational(length(xs) `div` 2))) xs, drop(ceiling(toRational(length(xs) `div` 2))) xs)
    else ((take (floor(toRational(length(xs) `div` 2))) xs, drop (floor(toRational(length(xs) `div` 2))) xs))
 main = do
 putStrLn(show (halve [1,2,3])) gives me [1],[2,3] instead of [1,2],[3]
 putStrLn(show (halve [])) gives me an error

这让我觉得mod中有一些截断操作或天花板操作。
我知道，对于putStrLn（show（halve []））的错误，解释器不知道数组是由什么组成的，但我怎样才能使它的类型无关紧要？我想这条线给我（[]，[]）。

This makes me think that there is some truncation going on in the mod operation or ceiling operation. I know that for the error with putStrLn(show (halve [])), the interpreter doesn't know what the array is made of, but how can I make it so that the type does not matter? I would like that line to give me ([],[]).

推荐答案

>

Have you tried the following:

halve :: [a] -> ([a], [a])
halve xs = let n = div (length xs + 1) 2
           in (take n xs, drop n xs)

-- or alternatively

halve :: [a] -> ([a], [a])
halve xs = splitAt (div (length xs + 1) 2) xs

div 函数向下舍去（楼层）正数商数并舍入（ceils）负商数：

The div function rounds down (floors) positive quotients and rounds up (ceils) negative quotients:

div  5 2 =>  2
div -5 2 => -2
div  6 2 =>  3
div -6 2 => -3

halve 你会期望它：

halve []     => ([], [])
halve [1..5] => ([1,2,3], [4,5])
halve [1..6] => ([1,2,3], [4,5,6])

我不看看为什么你需要一个 if then else 分支。您的功能不必要的复杂。

I don't see why you need an if then else branch. Your function is unnecessarily complicated.

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