标签: parse-error

这是我的代码:

select_where_true :: (Double -> Bool) -> [Double] -> [Double]
select_where_true is_neg [a] = case [a] of
[] -> []
x:xs -> is_neg x 
             |(is_neg x) == False = []
             |(is_neg x) == True = x ++ (select_where_true is_neg xs)


is_neg :: Double -> Bool
is_neg x = x < 0

这是错误消息:

[1 of 1] Compiling Main             ( test.hs, interpreted )

test.hs:5:18: parse error on input `|'
Failed, modules loaded: none.

有人喜欢告诉我我的代码有什么问题吗？

感谢任何能给我一些建议的人.

当我尝试编译它时,以下do块会抛出错误"输入`conn'上的解析错误".我尝试过if-then-else语句的许多不同配置无济于事.在添加条件之前数据库逻辑工作,所以没有问题.我在其他地方有太多行吗？有没有办法解决这个问题而不完全修改逻辑？

main = do
   contents <- BL.getContents
   let myData = decode contents :: Maybe Data
   if maybe True (\x -> result x /= "success") myData
      then error ("JSON download failed")
      else let myTrades = process myData
         conn <- connectSqlite3 "trades.db"
         insert <- DB.prepare conn "INSERT INTO trades VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?);"
         DB.executeMany insert $ map (\xs -> map DB.toSql xs) myTrades
         DB.commit conn
         DB.disconnect conn

我试图编写一个带有列表对的函数,并交换对元素

inverse :: [(a,b)] -> [(b,a)]
inverse [] = []
inverse (x,y):xs = (y:x): inverse xs

我通过Prelude加载了这个函数,它给了我以下错误:

mydefs.hs:11:1:模式中的解析错误:反向

这是第11行, inverse (x,y):xs = (y:x): inverse xs

我正在写一个程序来帮助我的弟弟学习加法.我没有编写IO程序的经验,而且我遇到了这个解析错误:

MyCode.hs:6:25:
    Parse error in pattern: show
    Possibly caused by a missing 'do'?

代码:

mathExercise times (a,b) = 
    if times<=0
    then return ()
    else do let x = randInt a
            let y = randInt b 
            putStr (show x ++ " + "++ show y++ " = ")
            ans <- getInt
            if (ans==x+y)
            then do print True
                    mathExercise (times-1) (a,b)
            else do print False

我有以下代码,旨在从表达式的评估中取出重复,m并替换开头的双重se

exec (Assign s e) m = assign s (eval e m) m
where assign _ _ [] = error ("undef var " ++ s)
    assign s v (x:xs)
        | fst x == s = if sameKind v (fst x)
                        then (fst x,v):xs
                        else error "type error in assign"
        | otherwise = x:(assign s v xs)
            where sameKind (VInt a) (VInt b) = True
                sameKind (VBool a) (VBool b) = True
                sameKind _ _ = False …

我的Web.config文件突然出现以下错误,我不明白这意味着什么:

分析器错误消息:属性"connectionStringName"缺失或为空.

Line 24:       <providers>
Line 25:         <clear />
Line 26:        <add name="SMDPortalMembershipProvider" type="SMDPortalMembershipProvider" />
Line 27:       </providers>
Line 28:     </membership>

源文件:c:\ inetpub\wwwroot\web.config行:26

版本信息:Microsoft .NET Framework版本:4.0.30319; ASP.NET版本:4.0.30319.272

这是我的配置文件:

<?xml version="1.0"?>
<!--
  For more information on how to configure your ASP.NET application, please visit
  http://go.microsoft.com/fwlink/?LinkId=169433
  -->
<configuration>
  <connectionStrings>
    <add name="ApplicationServices"
      connectionString="data source=.\SQLEXPRESS;Integrated Security=SSPI;AttachDBFilename=|DataDirectory|\aspnetdb.mdf;User Instance=true"
      providerName="System.Data.SqlClient"/>
  </connectionStrings>
  <system.web>
    <compilation debug="true" targetFramework="4.0">
      <assemblies>
        <add assembly="UODOTNET, Version=2.2.5.7444,
          Culture=neutral, PublicKeyToken=335F3FBD4BE82339"/>
        <add assembly="System.Core, Version=4.0.0.0,
          Culture=neutral, PublicKeyToken=B77A5C561934E089"/>
      </assemblies>
    </compilation>
    <authentication mode="Forms">
      <forms loginUrl="Default.aspx" timeout="2880" /> …

我刚刚开始使用Haskell,我无法理解一件事 - 为什么这样的东西不起作用？

fun f = f * f
main = 
    do 
        foo <- getLine 
        bar <- getLine
        print (fun 4)

我一直收到来自ghc的"输入`print'的解析错误".但对我来说更神秘的是它在移除foo < - getLine和bar < - getLine后起作用.有任何想法吗？

下面的代码被编程来决定Haskell中的纸岩和剪刀的结果,但是终端给出了错误

data Move = Paper | Rock | Scissors  
 deriving (Eq, Show)
data Result = Win | Draw | Loose 
  deriving (Eq, Show)


beats :: Move -> Move
beats move = case move of 
  Paper -> Scissors
  Rock  -> Paper
  Scissors -> Rock

score :: Move -> Move -> Result
score this_move opposing_move
  | this_move == beats opposing_move = Win
  | this_move == opposing_move       = Draw
  | otherwise = Loose

这是来自终端的错误消息

[1 of 1] Compiling Main             ( test.hs, interpreted ) …

我有一个语法显然是错误的,因为解析一个简单的文件会产生奇怪的错误信息.

我尽可能地简化了以下语法而没有改变错误(如果你删除'this'了ANTLRWorks的树gui输出,那么int样本文件的标记会有不同的颜色,尽管结构看起来是相同的).

grammar DepClsJ_no_java_debug;

module   : ( methodDecl )* ;

methodDecl   : pathType Identifier '()' block ;

pathType   : Identifier | 'this' ;

block   : '{'
    ( localDecl ';'  )*
    ( statement  )*
    ( expr  )?
    '}'   ;

localDecl   : pathType Identifier ( '=' expr )?;

statement   : block | expr ';' ;

expr   : dotExpr ( '=' dotExpr  )* ;    dotExpr   : Identifier ( '.' Identifier )* ;

Identifier   : ('a'..'z'|'A'..'Z'|'_') ('a'..'z'|'A'..'Z'|'0'..'9'|'_')* ;

演示代码:

void …

我是Haskell的初学者,想要开始解决它的问题,所以我试图解决第一个SPOJ问题(问题代码:TEST)."问题"是读取行并打印它们直到"42"出现.

main = do input <- getLine
          if input == "42" then putStr "" 
          else do putStrLn input
                  main

我的解决方案非常简单,但输入'if'时出现解析错误.当我在开头和else语句中将'main'更改为'main2'时,一切正常.为什么'if'上有解析错误？