Question

我有以下代码，它实现了筛子或Eratosthenes：

primeSieve :: Int -> [Int] -- Returns a list of primes up to upperbound
primeSieve upperbound = filter[2..upperbound] 
  where filter ls indx = 
    let divisor = (ls!!indx)
    let filtered = [x | x <- ls, x `mod` divisor /= 0]
    if divisor*divisor >= last filtered
      then filtered
      else filter filtered (indx+1)

我在第4行遇到了一个解析错误，“可能是错误的缩进或括号不匹配”。

这是为什么？

Answer 1

我相信你想用{x}来写// Getting the absolute root path define('ROOT', dirname(__FILE__) . DIRECTORY_SEPARATOR); include_once("App/Core/Config/config.php"); include_once(CORE_PATH . DIRECTORY_SEPARATOR . "Include/autoloader.php"); //echo ('location: http://' . $_SERVER['HTTP_HOST'] . $_SERVER['PHP_SELF'] . "?comp=login_controller/noAuth"); if (!isset($_GET['comp'])) { header('location: http://' . $_SERVER['HTTP_HOST'] . $_SERVER['PHP_SELF'] . "?comp=login_controller/noAuth"); } $app = new Application(); $app->run();。您可以在filter之后添加do。但是，这段代码是纯粹的（即非monadic），我会推荐这种语法：

filter ls indx =

...但是您的代码给了我以下错误：

primeSieve :: Int -> [Int] -- Returns a list of primes up to upperbound
primeSieve upperbound = filter [2..upperbound]
  where
    filter ls indx =
        let divisor = (ls!!indx)
            filtered = [x | x <- ls, x `mod` divisor /= 0]
        in if divisor*divisor >= last filtered
          then filtered
          else filter filtered (indx+1)

我认为您打算将<file>:13:25: Couldn't match expected type `[Int]' with actual type `Int -> [Int]' Probable cause: `filter' is applied to too few arguments In the expression: filter [2 .. upperbound] In an equation for `primeSieve': primeSieve upperbound = filter [2 .. upperbound] where filter ls indx = let ... in if divisor * divisor >= last filtered then filtered else filter filtered (indx + 1) Failed, modules loaded: none.作为第二个参数传递给0：

filter

Answer 2

问题是filter的定义的缩进比它应该的小。重要的是

where filter ls indx = 
  let divisor = (ls!!indx)

其中let在上面一行的filter之前开始4个字符。这会触发语法错误。要解决此问题，您可以缩进filter定义

where filter ls indx = 
        let divisor = (ls!!indx)

更常见的格式是

where
  filter ls indx = 
    let divisor = (ls!!indx)

因为这样你就不会缩进太多了。

您可以在haskell wiki上找到有关缩进的更多信息，其中包含有关此类错误的一些很好的可视化示例。

Haskell：嵌套where子句中的Parse Error

2 个答案: