读取字符串并测试它是否是数字

发布于 2024-11-08 17:45:11 字数 195 浏览 3 评论 0原文

为什么这段代码不起作用？如果字符串是数字，我想返回 Bool。

isNumber = do
    n <- getLine
    let val = case reads n of
                ((v,_):_) -> True
                _ -> False

原文

Why this code doesn't work ? I would like to return Bool if string is a number.

isNumber = do
    n <- getLine
    let val = case reads n of
                ((v,_):_) -> True
                _ -> False

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勿忘初心 2024-11-15 17:45:11

首先，您有一个语法错误：

A.hs:3:5:
    The last statement in a 'do' construct must be an expression:
    let val
          = case reads n of {
              ((v, _) : _) -> True
              _ -> False }

因为您的函数尚未返回值。修复：

isNumber = do
    n <- getLine
    let val = case reads n of
                ((v,_):_) -> True
                _         -> False
    return val

现在它在语法上是正确的，但有一个类型错误：

A.hs:3:20:
    Ambiguous type variable `a0' in the constraint:
      (Read a0) arising from a use of `reads'
    Probable fix: add a type signature that fixes these type variable(s)

为什么？因为read已经超载了。因此编译器不知道您要读取的内容。在本例中，您正在尝试读取一个数字。假设是一个Integer：

isNumber :: IO Bool
isNumber = do
    n <- getLine
    let val = case (reads :: ReadS Integer) n of
                ((v,_):_) -> True
                _         -> False
    return val

尽管如此，它并不是很惯用。让我们将 IO 与纯代码分开，如果成功的话，实际上返回解析的数字：

readNumber :: String -> Maybe Integer
readNumber s = case reads s of
            ((v,_):_) -> Just v
            _         -> Nothing

getNumber :: IO (Maybe Integer)
getNumber = do
    s <- getLine
    return (readNumber s)

测试：

*Main> getNumber 
123
Just 123
*Main> getNumber 
dons
Nothing

所以我们已经清理了解析，并将 IO 与解析分开，这意味着您可以单独测试解析器，并添加类型信息记录您的设计。

Firstly, you have a syntax error:

A.hs:3:5:
    The last statement in a 'do' construct must be an expression:
    let val
          = case reads n of {
              ((v, _) : _) -> True
              _ -> False }

Because your function doesn't return a value yet. Fixing that:

isNumber = do
    n <- getLine
    let val = case reads n of
                ((v,_):_) -> True
                _         -> False
    return val

Now it is syntactically correct, but it has a type error:

A.hs:3:20:
    Ambiguous type variable `a0' in the constraint:
      (Read a0) arising from a use of `reads'
    Probable fix: add a type signature that fixes these type variable(s)

Why? Because read is overloaded. So the compiler doesn't know what you're trying to read. In this case, you're trying to read a number. Let's say, an Integer:

isNumber :: IO Bool
isNumber = do
    n <- getLine
    let val = case (reads :: ReadS Integer) n of
                ((v,_):_) -> True
                _         -> False
    return val

Still, its not really idiomatic. Let's separate the IO from the pure code, and actually return the parsed number, if it succeeds:

readNumber :: String -> Maybe Integer
readNumber s = case reads s of
            ((v,_):_) -> Just v
            _         -> Nothing

getNumber :: IO (Maybe Integer)
getNumber = do
    s <- getLine
    return (readNumber s)

Testing:

*Main> getNumber 
123
Just 123
*Main> getNumber 
dons
Nothing

So we've cleaned up the parsing, and separted IO from parsing, meaning you can test your parser in isolation, and added type information to document your design.

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失与倦＂ 2024-11-15 17:45:11

添加return val或简单地编写return $ case ...。 do ... 中的最后一个语句必须是表达式。在您的特定情况下，它必须是 IO Bool 类型的表达式，因此您需要使用 return 函数将值提升到 IO monad 中。您还需要显式指定 v 的类型（为此，您可能需要 ScopedTypeVariables GHC 扩展。）

编写一个单独的类型 < 的纯函数也是一个好主意。代码>字符串-> Bool 并在不纯的 IO 代码中使用它。