***例外：Prelude.read：Haskell中不解析 - 解析，表達式和遞歸

Question

這部分代碼應該讀取兩個或更多數字（省略主io函數），然後用「+」來給出總和。理性被使用，因爲後來我會做乘法和其他操作。

data Expression = Number Rational 
       | Add (Expression)(Expression) 
       deriving(Show,Eq) 

solve :: Expression -> Expression 
solve (Add (Number x) (Number y)) = Number (x + y) 

parse :: [String] -> [Expression] -> Double 
parse ("+":s)(a:b:xs) = parse s (solve (Add a b):xs) 
parse [] (answer:xs) = fromRational (toRational (read (show answer)::Float)) 
parse (x:xs) (y) = parse (xs) ((Number (toRational (read x::Float))):y) 

（第二）錯誤是用解析功能無法處理

*Main> parse ["1","2","+"] [Number 3] 

*** Exception: Prelude.read: no parse 

我已經看過了Data.Ratio頁面和網頁此解決方案上，但還沒有找到它，會感謝一些幫助。謝謝，

CSJC

Answer 1

第一個方程，

parse ("+":s)(a:b:xs) = parse (s)((solve (Add (Number a) (Number b))):xs) 

應該

parse ("+":s)(a:b:xs) = parse (s)((solve (Add a b)):xs) 

，因爲每個類型簽名，a和b已經是Expression秒。

或者，在與第二和第三方程線，所述類型更改爲

parse :: [String] -> [Rational] -> Double 

和第一方程改變爲固定代碼的

parse ("+":s)(a:b:xs) = parse s ((a + b):xs) 

兩種可能的方法（有是更有問題的部分）：

-- Complete solve to handle all cases 
solve :: Expression -> Expression 
solve expr@(Number _) = expr 
solve (Add (Number x) (Number y)) = Number (x + y) 
solve (Add x y) = solve (Add (solve x) (solve y)) 

-- Convert an Expression to Double 
toDouble :: Expression -> Double 
toDouble (Number x) = fromRational x 
toDouble e = toDouble (solve e) 

-- parse using a stack of `Expression`s 
parse :: [String] -> [Expression] -> Double 
parse ("+":s) (a:b:xs) = parse s ((solve (Add a b)):xs) 
parse [] (answer:_) = toDouble answer 
parse (x:xs) ys = parse xs (Number (toRational (read x :: Double)) : ys) 
parse _ _ = 0 

-- parse using a stack of `Rational`s 
parseR :: [String] -> [Rational] -> Double 
parseR ("+":s) (a:b:xs) = parseR s (a+b : xs) 
parseR [] (answer:xs) = fromRational answer 
parseR (x:xs) y = parseR xs ((toRational (read x::Double)):y) 
parseR _ _ = 0 

後者比較謹慎，因爲最終生成了Double，所以使用Rational作爲堆棧沒有真正的意義。

在您的parse代碼，第三個方程省去了通過Number構造一個Rational到Expression的轉換，但在其他方面正常。第二個公式，但是，含有不同類型的問題：

parse [] (answer:xs) = fromRational (toRational (read (show answer)::Float)) 

如果answer或者是一個Expression或Rational，show answer不能被解析爲一個Float，這樣將導致一個運行時間錯誤，通過所例示您編輯：

（第二）錯誤是用解析功能無法處理

*Main> parse ["1","2","+"] [Number 3] 
*** Exception: Prelude.read: no parse 

在其中使用所述第二方程點，堆疊在所述第一元件（answer）是Number (3 % 1)，和show (Number (3 % 1))是"Number (3 % 1)"，其不是String該read可以作爲一個Float解析。