Haskell解析前缀评估 [英] Haskell parsing prefix evaluation
问题描述
import Data.Char(isDigit)
data Ast = V Int | Neg Ast | A Ast Ast | M表示方程式(Show,Eq)
parseE(+:xs)= let(e1,r1)= parseE xs; (e2,r2)= parseE r1 in(A e1 e2,r2)
parseE(*:xs)= let(e1,r1)= parseE xs; (e2,r2)= parseE r1 in(M e1 e2,r2)
parseE( - :xs)= let(a,r)= parseE r in(N a,r)
parseE (a,r)
parseE(x:xs)=(V(read x :: Int),xs)((:xs)= let(a,):r)= parseE r
eval xs = parseE xs
当我的输入如下所示: * + 1 2 * 3 + 7 - 2
我希望输出为: ((1 + 2)* 3)*(3 *(7-2))
应显示 45
当我在haskell中加载我的文件时,出现此错误:
:loadnewf。 hs
[1的1]编译主(newf.hs,解释)
newf.hs:6:44:错误:数据构造函数不在范围内:N :: Ast - > ; Ast
|
6 | parseE( - :xs)= let(a,r)= parseE r in(N a,r)
| ^
失败,加载了0个模块。
错误信息表明数据构造函数 N
不在范围内。数据构造函数是 data
语句右侧的内容。在你的例子中, V
, Neg
, A
和 M
是数据构造函数。 不在范围内的意思是没有定义它在哪里使用。
看起来你写了 N
where你打算写 Neg
,反之亦然。将您的 data
语句修改为:
data Ast = V Int | N Ast | A Ast Ast | M Ast Ast派生(Show,Eq)
允许程序编译。
程序中仍然存在一些错误。例如,以下内容会陷入循环:
> (A(V 1)(V 2))(M(V 3)(A(V 7)(N - 冻结)
因为声明:
<$ p $ (a,r)= parseE r in(N a,r)
它引入了一个无意的递归定义 - 您正在定义 r
作为调用 parseE r
的结果,这会导致无限循环。在尝试处理括号的情况下,您也遇到了类似的问题。
I'm struggling with this code.
import Data.Char (isDigit)
data Ast = V Int | Neg Ast | A Ast Ast | M Ast Ast deriving (Show,Eq)
parseE ("+":xs) = let (e1,r1) = parseE xs; (e2,r2) = parseE r1 in (A e1 e2, r2)
parseE ("*":xs) = let (e1,r1) = parseE xs; (e2,r2) = parseE r1 in (M e1 e2, r2)
parseE ("-":xs) = let (a,r) = parseE r in (N a, r)
parseE ("(":xs) = let (a,")":r) = parseE r in (a,r)
parseE (x:xs) = (V (read x :: Int), xs)
eval xs = parseE xs
When my input is something like: * + 1 2 * 3 + 7 - 2
I want the ouput to be: ((1+2)*3)*(3*(7-2))
which should show 45
When I load my file in haskell, I get this error :
:load "newf.hs"
[1 of 1] Compiling Main ( newf.hs, interpreted )
newf.hs:6:44: error: Data constructor not in scope: N :: Ast -> Ast
|
6 | parseE ("-":xs) = let (a,r) = parseE r in (N a, r)
| ^
Failed, 0 modules loaded.
The error message says that the data constructor N
is not in scope. Data constructors are the things on the right-hand side of data
statements. In your example, V
, Neg
, A
, and M
are data constructors. "Not in scope" means "not defined where it was used".
It looks like you wrote N
where you meant to write Neg
, or vice versa. Fixing your data
statement to read:
data Ast = V Int | N Ast | A Ast Ast | M Ast Ast deriving (Show,Eq)
allows the program to compile.
There are still a few bugs in the program. For example, the following gets stuck in a loop:
> parseE (words "* + 1 2 * 3 + 7 - 2")
(M (A (V 1) (V 2)) (M (V 3) (A (V 7) (N -- freezes
because of the statement:
let (a,r) = parseE r in (N a, r)
which introduces an unintended recursive definition -- you're defining r
as the result of calling parseE r
, which causes an infinite loop. You have a similar problem in the case that tries to handle parentheses.
这篇关于Haskell解析前缀评估的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!