Question

我提供了this question的答案，并且想到了使用Cont monad的想法。我对Haskell不够了解，无法解释为什么该程序不起作用

import Control.Monad.Cont

fib1 n = runCont (slow n) id
  where
    slow 0 = return 0
    slow 1 = return 1
    slow n = do
      a <- slow (n - 1)
      b <- slow (n - 2)
      return a + b

main = do
  putStrLn $ show $ fib1 10

错误-

main.hs:10:18: error:
    • Occurs check: cannot construct the infinite type: a2 ~ m a2
    • In the second argument of ‘(+)’, namely ‘b’
      In a stmt of a 'do' block: return a + b
      In the expression:
        do a <- slow (n - 1)
           b <- slow (n - 2)
           return a + b
    • Relevant bindings include
        b :: a2 (bound at main.hs:9:7)
        a :: a2 (bound at main.hs:8:7)
        slow :: a1 -> m a2 (bound at main.hs:5:5)
   |
10 |       return a + b
   |

但这对我来说没有意义。为什么有a2和m a2？我期望a和b属于同一类型。

这让我很烦，因为同一程序在JavaScript中工作得很好。也许Haskell需要一个类型提示？

const runCont = m => k =>
  m (k)

const _return = x =>
  k => k (x)
  
const slow = n =>
  n < 2
    ? _return (n)
    : slow (n - 1) (a =>
      slow (n - 2) (b =>
      _return (a + b)))
      
const fib = n =>
  runCont (slow(n)) (console.log)
  
fib (10) // 55

Answer 1

return a + b解析为(return a) + b，而您想要return (a + b)。请记住，函数应用程序的绑定比任何infix运算符都更紧密。

（写return $ a + b也很常见，这等同于同一件事）

如何在递归函数中使用Control.Monad.Cont？

1 个答案: