Question

quicksort :: (Ord a) => [a] -> [a]
quicksort [] = []
quicksort (x:xs) =
    let smallerOrEqual = [a | a <- xs, a <= x]
        larger = [a | a <- xs, a > x]
    in quicksort smallerOrEqual ++ [x] += larger

main = do
    a = [ 5, 1, 9, 4, 6, 7, 3]
    print quicksort a

我收到了这个错误：

[1 of 1] Compiling Main             ( quicksort.hs, quicksort.o )

quicksort.hs:10:11: parse error on input `='

真的不明白为什么。双重检查此代码几次。仍然令人困惑。需要帮忙。谢谢！

Answer 1

在do个区块内，绑定必须以let关键字开头，因此该行应为let a = [ 5, 1, 9, 4, 6, 7, 3]。

但是还有一些其他错误。您似乎错误输入了++运算符，而您忘记对列表的较大部分进行排序。您的打印声明无法正常工作，因为print quicksort a被解析为将函数quickSort和值a传递给print。您可以使用括号或函数应用程序运算符（$）轻松解决此问题。这段代码完美无缺：

quicksort :: (Ord a) => [a] -> [a]
quicksort [] = []
quicksort (x:xs) =
    let smallerOrEqual = [a | a <- xs, a <= x]
        larger = [a | a <- xs, a > x]
    in quicksort smallerOrEqual ++ [x] ++ quickSort larger

main = do
    let a = [ 5, 1, 9, 4, 6, 7, 3]
    print $ quicksort a

Answer 2

您忘记了let，

quicksort :: (Ord a) => [a] -> [a]
quicksort [] = []
quicksort (x:xs) =
    let smallerOrEqual = [a | a <- xs, a <= x]
        larger = [a | a <- xs, a > x]
    in quicksort smallerOrEqual ++ [x] ++ quicksort larger -- ++ not +=

main = do
    let a = [ 5, 1, 9, 4, 6, 7, 3]
    print (quicksort a) -- The parens are needed too

我也解决了其他一些错误。

值得注意的是，这个快速排序并不合适，因此它并不是真正的快速排序，尽管很漂亮。

输入“=”上的haskell解析错误

2 个答案: