相关疑难解决方法(0)

我期待找到一个功能

integer :: Stream s m Char => ParsecT s u m Integer

或者甚至是

natural :: Stream s m Char => ParsecT s u m Integer

在标准库中,但我没有找到一个.

将纯自然数直接解析为Integer？的标准方法是什么？

我有以下类型检查:

p_int = liftA read (many (char ' ') *> many1 digit <* many (char ' '))

现在,正如函数名称所暗示的那样,我希望它能给我一个Int.但如果我这样做:

p_int = liftA read (many (char ' ') *> many1 digit <* many (char ' ')) :: Int

我收到这种类型的错误:

Couldn't match expected type `Int' with actual type `f0 b0'
In the return type of a call of `liftA'
In the expression:
    liftA read (many (char ' ') *> many1 digit <* many (char ' ')) ::
      Int
In an equation for `p_int': …