Question

create_subseq size xs = 
  if (length xs) == size
  then [ [ x | x <- s] | s <- xs] 
  else [ [ i | i <- subxs] | subxs <- (take size xs)] ++ create_subseq size (tail xs)

我一直在尝试使用下面的行在ghci中运行此代码但是我得到下面的错误

create_subseq 3 [1,2,3,4]

No instance for (Num [t0]) arising from the literal `1'
Possible fix: add an instance declaration for (Num [t0])
In the expression: 1
In the second argument of `create_subseq', namely `[1, 2, 3, 4]'
In the expression: create_subseq 3 [1, 2, 3, 4]

Answer 1

如果您查看create_subseq中ghci的类型，就会明白这一点：

*Main> :t create_subseq
create_subseq :: Int -> [[t]] -> [[t]]

换句话说，第二个参数应该是某个列表的列表，但是您只传递了一个数字列表。 ghci然后尝试将数字解释为列表，但它不能，因此错误消息。

如果您想知道为什么create_subseq会列出一系列列表，请查看您编写的列表推导。他们每个人都假设原始参数xs是一个列表，当你查看列表xs中的一个元素时，你有另一个列表，你可以写一个嵌套列表理解。

Answer 2

这是我的代码，我开始弄清楚haskell。到目前为止我喜欢它，我希望继续学习。

-- Program will read a file with a 1000 digit number. The goal of the program
-- is to find a sequence within that number that is 13 digits long and has
-- the largest product. This program is not safe to run on 32 bit machines
-- you will run into number overflow and get inaccurate results

import System.Environment
import Data.Char

create_sub_seq :: Int->[Int]->[[Int]]
create_sub_seq n xs = if length xs == n
              then [xs]
              else [[i | i <- take n xs]]++create_sub_seq n (tail xs)

prod :: [Int]->Int
prod xs = foldl (*) 1 xs 

getIntArray :: [Char] -> [Int]
getIntArray xs = [digitToInt i |i <- xs, not(i=='\n')]


max_sub_seq :: ([Int]->Int)->[[Int]]->[Int]
max_sub_seq  f [] = []
max_sub_seq  f (xs:xxs) = if (f b) > (f xs) 
                  then b
                  else xs
                  where b=max_sub_seq f xxs

main = do [f] <- getArgs
      s <- readFile f

--    print( create_sub_seq 4 (getIntArray s))
      print( max_sub_seq prod (create_sub_seq 13 (getIntArray s)) )
      print( prod ( max_sub_seq prod (create_sub_seq 13 (getIntArray s)) ) )

没有（Num [t0]）的实例来自字面上的'1＆＃39;

2 个答案: