我正在提供一个答案这个问题 https://stackoverflow.com/q/54843670/633183我想到了一个可以使用的想法Cont
单子。我不太了解 Haskell,无法解释为什么这个程序不起作用
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
Error -
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。为什么我有a2
and m a2
?我期待着a
and b
为同一类型。
这让我很烦恼,因为同一个程序在 JavaScript 中运行得很好。也许 Haskell 需要类型提示?
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
return a + b
解析为(return a) + b
,而你想要return (a + b)
。请记住,函数应用程序比任何中缀运算符绑定得更紧密。
(也很常见写return $ a + b
,这相当于同一件事)
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