我正在寻找fibonacci级数的和。这就是我被困的地方:
fibs = 0 : 1 : zipWith (+) fibs (tail fibs)
main = do putStrLn "Enter a number:"
num <- readLn
foldr (+) 0 (take num fibs)
错误为:
No instance for (Num (IO t0))
arising from the literal `0'
Possible fix: add an instance declaration for (Num (IO t0))
In the second argument of `foldr', namely `0'
In the expression: foldr (+) 0 (take num fibs)
In the expression:
do { putStrLn "Enter a number:";
num <- readLn;
foldr (+) 0 (take num fibs) }
我到底哪里出了问题?
您可能想要print
结果:
main = do putStrLn "Enter a number:"
num <- readLn
print $ foldr (+) 0 (take num fibs)
出现错误消息的原因是do
块中的每个语句都必须属于同一个monad。在main
的情况下,这就是IO
。然而,这里foldr
的结果是一个数字,而不是IO
的动作。
错误消息令人困惑,因为GHC明智地得出结论,IO
操作必须是数字,这当然是无稽之谈。
当您遇到一个令人困惑的类型错误时,通常可以为所涉及的一些表达式添加一些类型注释,向GHC解释您期望的类型。这通常会给你更好的错误信息从GHC返回。
例如,如果在带有foldr
的行的末尾添加:: Integer
,则会得到以下消息:
Couldn't match expected type `IO b0' with actual type `Integer'
In a stmt of a 'do' block: foldr (+) 0 (take num fibs) :: Integer
In the expression:
do { putStrLn "Enter a number:";
num <- readLn;
foldr (+) 0 (take num fibs) :: Integer }
In an equation for `main':
main
= do { putStrLn "Enter a number:";
num <- readLn;
foldr (+) 0 (take num fibs) :: Integer }
在这里更容易看出问题。GHC期望CCD_ 10类型的语句,你给了它一个Integer
。