Haskell：模式中的解析错误：a - 1 可能是由缺少"do"引起的？

Question

在尝试在Haskell中实现著名的Ackermann函数以测试所承诺的运行时间差异是否实际上是可测量的时候，我偶然发现了这个美丽的错误描述:

Parse error in pattern: a - 1 
Possibly caused by a missing 'do'?

我知道这个解析错误很常见，我不知道为什么会这样。除此之外，我的代码(见下文)应该没问题。

我代码:

main :: IO ()
main = do 
print "Please enter first operand: "
input <- getLine
let n = read input
print "Please enter second operand: "
input <- getLine
let m = read input
let r = ak(n,m)
print(n++" ackermann "++m++" = "++show r)

ak(a,b) = do
if a == 0 
then return (b+1)
else if b == 0 
then return ak(a-1, 1)
else 
s1 <- ak(a-1, b)
s2 <- ak(a-1, s1)
return s2

Answer 1

不要用do和return代替ak。do块用于一元计算。您可以使用:

ak 0 b = b + 1
ak a 0 = ak (a-1) 1
ak a b = ak (a-1) (ak a (b-1))

那么你可以实现main为:

main :: IO ()
main = do 
print "Please enter first operand: "
n <- readLn
print "Please enter second operand: "
m <- readLn
putStrLn (show n ++ " ackermann " ++ show m ++ " = " ++ show (ak n m))