Haskell:使用foldr在函数中输入不匹配

Question

我不明白为什么这些函数有错误:

countEqualPairs:: Eq a => [a] -> Int
countEqualPairs (s:ss) = foldr test s ss

test :: Eq a => a -> a -> Int
test s c = if (c == s) then 1 else 0

错误信息:

Could not deduce (a ~ Int)
from the context (Eq a)
  bound by the type signature for
            countEqualPairs :: Eq a => [a] -> Int
  at Blatt06.hs:30:20-37
 `a' is a rigid type variable bound by
     the type signature for countEqualPairs :: Eq a => [a] -> Int
     at Blatt06.hs:30:20
Relevant bindings include
 ss :: [a] (bound at Blatt06.hs:31:20)
  s :: a (bound at Blatt06.hs:31:18)
 countEqualPairs :: [a] -> Int (bound at Blatt06.hs:31:1)
 In the second argument of `foldr', namely `s'
 In the expression: foldr test s ss

有谁可以解释,我错了什么？

谢谢!

Answer 1

foldr :: (a -> b -> b) -> b -> [a] -> b可以在Hoogle上找到该表单的功能a -> b -> b,但您的test功能看起来很像a -> a -> b.

正如Carsten所提到的b,你的函数甚至与b累加器不匹配,并且在使用时可以从GHCi中检索信息:t foldr(这可以用于任何函数).您尝试做的事情的替代方案可能是(如果我做对了,你试着计算双打的数量):

countEqualPairs [] = 0
countEqualPairs (s:ss) = if (s `elem` ss) then 1 + x else x
                         where x = countEqualPairs ss