Question

This portion of code should read in two or more numbers (main io function omitted), then a "+" to give the sum. Rationals are used because later i will do multiplications and such other operations.

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)

(二) 差错是由于无法处理 par子功能。

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

*** Exception: Prelude.read: no parse

我研究了数据。这一解决办法的网页和网页上,有点发现,希望得到一些帮助。感谢

CSJC

Answer 1

第一个方程式,

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

应当:

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

自每一类型签名以来,a和b 页: 1

<>Or/strong>,根据第二和第三类,将类型改为

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

并改变第一个方程式

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

两种确定守则的可能方法(更多有问题部分):

-- 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

后者相当细致,因为最终产生了<代码>Double<>/code>,因此没有使用<代码>Rational的实际点。

在您的<代码>parse上,第三种方程式将<代码>Rational转换成Expression通过Number的构造者,但以其他方式予以罚款。然而,第二个方程式含有不同类型的问题:

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

如果answer要么是Expression,要么是Rational,showwer不能作为Float加以统一,从而产生错误,如你的编辑所示:

(二) 差错是由于无法处理 par子功能。

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

在使用第二个方程式时,中值的第一个要素(answer)为Number (3 % 1)和show (Number (3 % 1)为Number (3 % 1)",后者不是String。可查阅。

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