Haskell：模式中的解析错误：A -1可能是由缺失＆＃x27; do＆＃x27引起的？

发布于 2025-02-09 08:56:07 字数 808 浏览 1 评论 0原文

试图实施著名的 ackermann函数在Haskell中实际上是可衡量的，我偶然发现了错误描述的美丽：

 模式中解析错误：A -1 
   可能是由于缺失的“做”引起的？

我知道这个解析错误很常见，我不知道为什么这样做。除此之外，我的代码（见下文）应该很好。

我的代码：

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

原文

While trying to implement the famous Ackermann function in Haskell to test whether the promised difference in running time is actually measurable, I stumbled across this beauty of an error description:

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

I know that this parsing error is appearing pretty commonly and I have no idea why it does so.
Other than that, my code (see below) should be fine.

My code:

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

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日裸衫吸 2025-02-16 08:56:07

不要使用do和返回 ak。 do块用于 monadic 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))

然后可以实现main as：

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))

Don't use do and return for ak. do blocks are used for monadic computations. You can work with:

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

Then you can implement main as:

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))

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