haskell - 为什么这会给我 "is a rigid type variable bound by"错误

Question

heist :: (Num n) => [n] -> [n] -> n -> n
-- heist [] [] _ = 0
heist w v maxw = func w v i j where
    i = length w
    j = maxw  
func :: (Num n) => [n] -> [n] -> n -> n -> n 
func _ _ 0 0 = 0

上面的代码给了我:

Heist.hs:15:27:    Could not deduce (n ~ Int)    from the context (Num n)      bound by the type signature for                 heist :: Num n => [n] -> [n] -> n -> n      at Heist.hs:(15,1)-(17,16)      `n' is a rigid type variable bound by          the type signature for heist :: Num n => [n] -> [n] -> n -> n          at Heist.hs:15:1    In the third argument of `func', namely `i'    In the expression: func w v i j    In an equation for `heist':        heist w v maxw          = func w v i j          where              i = length w              j = maxw

为什么会发生这种情况？

我正在一寸一寸地研究 Haskell 类型系统

Answer 1

length 始终返回一个 Int。通过将 i 传递给 func，您是在说 n 应该是 Int，但是 heist 希望 n 是通用的，因此会出现类型错误。