未定义的索引通知('Undefined index notice')

 尝试使用return $ BS.unpack headers而不是return (headers >>= BS.unpack) 。 
 
 或者如果headers是ByteStrings列表，请尝试return $ map BS.unpack headers头。 
 
 除了它碰巧类型检查（我假设BS.unpack headers工作）的事实，这里是一种思考事物的方式： 
 
 
 
 headers是纯粹的价值 
 
 
 BS.unpack是一个纯函数 
 
 
 headers >>= ...没有意义，因为>>=的LHS需要是monadic计算 
 
 
 ... >>= BS.unpack没有意义，因为>>=的RHS需要是一个产生monadic计算的函数 
 
 
 BS.unpack headers是我们想要返回的字符串，但它是纯值 
 
 
 因此，我们使用return来将纯值推广为monadic计算 
 
 
 更新： 
 
 以下代码显示如果fetchHeader具有类型IO [BS.ByteString] ，那么您的代码将键入check： 
 
import Data.ByteString.Char8 as BS

fetchHeader :: IO [BS.ByteString]   -- this works
-- fetchHeader :: IO BS.ByteString  -- this doesn't
fetchHeader = undefined

foo :: IO String
foo = do
  headers <- fetchHeader
  return $ headers >>= BS.unpack
 
 另一方面，如果您将其类型更改为IO BS.ByteString ，则会遇到您遇到的错误。 
 
 更新2 ： 
 
 有趣的是，当headers是ByteStrings的列表时，表达式headers >>= BS.unpack确实有意义并且相当于： 
 
concat $ map BS.unpack headers

Try using return $ BS.unpack headers instead of return (headers >>= BS.unpack).
 
Or try return $ map BS.unpack headers if headers is a list of ByteStrings.
 
Besides the fact that it happens to type check (and I'm assuming BS.unpack headers works), here's a way to think about things:
 
 
 
headers is a pure value
 
 
BS.unpack is a pure function
 
 
headers >>= ... doesn't make sense because the LHS of >>= needs to be a monadic computation
 
 
... >>= BS.unpack doesn't make sense because the RHS of >>= needs to be a function which produces a monadic computation
 
 
BS.unpack headers is the string we want to return, but it's a pure value
 
 
we therefore use return to promote the pure value to a monadic computation
 
 
Update:
 
The following code shows that if fetchHeader has type IO [BS.ByteString], then your code will type check:
 
import Data.ByteString.Char8 as BS

fetchHeader :: IO [BS.ByteString]   -- this works
-- fetchHeader :: IO BS.ByteString  -- this doesn't
fetchHeader = undefined

foo :: IO String
foo = do
  headers <- fetchHeader
  return $ headers >>= BS.unpack
 
On the other hand, if you change its type to IO BS.ByteString you get the error you encountered.
 
Update 2:
 
Interestingly enough, when headers is a list of ByteStrings, the expression headers >>= BS.unpack does make sense and is equivalent to:
 
concat $ map BS.unpack headers