Question

不知道为什么调试多次仍然有错误，即使显式声明类型 int，不知道b0，a0，b2来自哪里

 ghc -o hello main.hs
[1 of 1] Compiling Main             ( main.hs, main.o )

main.hs:92:10:
    Couldn't match expected type `(Int, b0)'
                with actual type `(Int, Int, Int)'
    In the first argument of `fst', namely `x'
    In the first argument of `(==)', namely `fst x'
    In the first argument of `(&&)', namely `fst x == a1'

main.hs:92:25:
    Couldn't match expected type `((Int, Int, Int) -> Int, b1)'
                with actual type `(a0, b2) -> b2'
    In the first argument of `fst', namely `snd'
    In the first argument of `(==)', namely `fst snd x'
    In the second argument of `(&&)', namely `fst snd x == b1'

main.hs:93:12:
    Couldn't match expected type `(a1, (Int, Int, Int) -> Int)'
                with actual type `(a2, b3) -> b3'
    In the first argument of `snd', namely `snd'
    In the expression: snd snd x
    In a stmt of a 'do' block:
      if fst x == a1 && fst snd x == b1 then snd snd x else 0

main.hs:97:10:
    Couldn't match expected type `(Int, b4)'
                with actual type `(Int, Int, Int)'
    In the first argument of `fst', namely `x'
    In the first argument of `(==)', namely `fst x'
    In the first argument of `(&&)', namely `fst x == a1'

main.hs:97:25:
    Couldn't match expected type `((Int, Int, Int) -> Int, b5)'
                with actual type `(a3, b6) -> b6'
    In the first argument of `fst', namely `snd'
    In the first argument of `(==)', namely `fst snd x'
    In the second argument of `(&&)', namely `fst snd x == b1'

main.hs:98:12:
    Couldn't match expected type `(a4, (Int, Int, Int) -> Int)'
                with actual type `(a5, b7) -> b7'
    In the first argument of `snd', namely `snd'
    In the expression: snd snd x
    In a stmt of a 'do' block:
      if fst x == a1 && fst snd x == b1 then
          snd snd x
      else
          firstlogic xs a1 b1

这段代码主要是使用生成的逻辑表作为函数来返回值输入两个参数后的逻辑代码：

comb0 :: [(Int, Int, Int)]
comb0 = do
   a <- [0,1,2]
   b <- [0,1,2]
   return (a, b, max a b)

firstlogic :: [(Int, Int, Int)] -> Int -> Int -> Int
firstlogic [] a1 b1 = 0
firstlogic [x] a1 b1 = do
  if fst x == a1 && fst snd x == b1
  then snd snd x
  else
    0
firstlogic (x:xs) a1 b1 = do
  if fst x == a1 && fst snd x == b1
  then snd snd x 
  else
    firstlogic xs a1 b1

main :: IO()
main = do 
  print firstlogic comb0 1 2

Answer 1

函数fst需要一对，而不是三元组。

fst :: (a, b) -> a

您需要一个不同的功能，fst3：

fst3 :: (a, b, c) -> a

您可以编写此功能，也可以使用模式匹配/切换表达式。

fst snd x的外观也没有意义，因为fst只接受一个参数，而x仍然有(Int, Int, Int)类型。不确定你在这里想要完成什么。尝试使用模式匹配或切换表达式。

这并不是详尽无遗的。代码中还有其他问题。

Answer 2

正如迪特里希已经指出你已经解决的主要问题，我认为我可以帮助你多一点。

这个将编译，我认为它做你想要的（虽然我不确定）：

comb0 :: [(Int, Int, Int)]
comb0 = do
   a <- [0,1,2]
   b <- [0,1,2]
   return (a, b, max a b)

firstlogic :: [(Int, Int, Int)] -> Int -> Int -> Int
firstlogic [] _ _ = 0
firstlogic ((x1,x2,x3):xs) a1 b1
  | x1==a1 && x2==b1 = x3
  | otherwise = firstlogic xs a1 b1


main :: IO ()
main = 
  print $ firstlogic comb0 1 2

我不确定它是否符合您的要求的原因是我不明白您要做的事情 - 首先您将[0,1,2]的每个组合与其自身相结合，添加最大值这两部分然后您在firstlogic中搜索某个组合并再次返回max

如果我是对的，这个功能完全相同，但不需要comb0，也不需要搜索：

firstlogic :: Int -> Int -> Int
firstlogic = max

（好的：它不是检查两个输入是否在[0,1,2] - 但你可以轻松添加）

无法匹配预期类型`（（Int，Int，Int） - ＆gt; Int，b5）＆＃39;

2 个答案: