Haskell中条件定义的递归给出错误（发生检查：无法构造无限类型:)

Question

我想定义一个条件函数并在Haskell（ghci 7.x）中使用递归，如：

partitions n k m = gfunct n k m
       where
       gfunct n k m
         | n == 0 && k == 0 = 1
         | n < m  || k == 0 = 0
         | otherwise        = gfunct (n-m,k-1,m+1) + gfunct (n,k,m+1)

我收到一个错误：

fib.hs:20:35:
Occurs check: cannot construct the infinite type:
  a1 ~ t6 -> (t5, t6, t7) -> a1
Relevant bindings include
  m :: (t5, t6, t7) (bound at fib.hs:17:23)
  k :: t6 (bound at fib.hs:17:21)
  n :: (t5, t6, t7) (bound at fib.hs:17:19)
  gfunct :: (t5, t6, t7) -> t6 -> (t5, t6, t7) -> a1
    (bound at fib.hs:17:12)
In the first argument of ‘(+)’, namely
  ‘gfunct (n - m, k - 1, m + 1)’
In the expression:
  gfunct (n - m, k - 1, m + 1) + gfunct (n, k, m + 1)

我究竟做错了什么？

Answer 1

一个功能

g a b c = 0

如果由三个值a ， b和c提供，则返回0 ; 但是一个功能

g (a,b,c) = 0

如果由一个值(a,b,c)提供，则返回0 。 这是价值的三倍 ， a ， b和c 。