haskell - let..in 中 where 子句的范围

Question

encode :: String -> String
encode xs =
  let s = normalize xs
   in case s of
        [] -> []
        _ -> encode' s row col
        where row = isqrt (length s)
              col = length s `div` row

这会给出错误“变量不在范围内:s​​”。看来 let..in 中的 where 子句无法看到之前定义的变量 s。

我该如何重写这个？

Answer 1

where 是在函数级别定义的，由于 s 没有在该级别定义，因此无法访问 s。您可以在 let ... in ... 级别的头部添加额外的声明:

encode :: String -> String
encode xs =
  let s = normalize xs
      <strong>row = isqrt (length s)</strong>
      <strong>col = length s `div` row</strong>
   in case s of
        [] -> []
        _ -> encode' s row col

或者将所有内容移至 where 子句:

encode :: String -> String
encode xs = case s of
        [] -> []
        _ -> encode' s row col
  where s = normalize xs
        ns = length s
        <strong>row = isqrt ns</strong>
        <strong>col = ns `div` row</strong>