Re: [Haskell-beginners] pipe and IO

22 Mar 2016

      Yield in your producer is of type DataFrameHkI3D -> Producer DataFrameHkl3D
IO ()

But you are putting an (IO DataFrameHkI3D) in as an argument.

You should be able to do something like this:

getDataFrameHkl3D :: DataFrameHkl3DH5SixsUhv -> Producer DataFrameHkl3D IO
()
getDataFrameHkl3D d = do
  n <- lift $ hkl_h5_len (h5mu d)
  frames <- forM [0..n] $ \i -> lift (getDataFrameHkl3D' d i)  -- ::
Producer DataFrameHkl3D IO [DataFrameHkl3D]
  forM_ frames yield

I have a feeling it may be more efficient to yield immediately on each loop
as such:

getDataFrameHkl3D :: DataFrameHkl3DH5SixsUhv -> Producer DataFrameHkl3D IO
()
getDataFrameHkl3D d = do
  n <- lift $ hkl_h5_len (h5mu d)
  forM_ [0..n] (\i -> lift (getDataFrameHkl3D' d i) >>= yield)

You may also substitutde liftIO for lift, if you wish.

On Tue, Mar 22, 2016 at 12:37 PM, PICCA Frederic-Emmanuel <
frederic-emmanuel.picca@synchrotron-soleil.fr> wrote:
...
Hello,
Still playing with the Pipe module
here the code I try to write
getDataFrameHkl3D' ::  DataFrameHkl3DH5SixsUhv -> Int -> IO DataFrameHkl3D
getDataFrameHkl3D' d i = do
  positions <- get_position (h5mu d) i
  return $ DataFrameHkl3D { df_n = i
                          , df_image = positions
                          }
getDataFrameHkl3D :: DataFrameHkl3DH5SixsUhv -> Producer DataFrameHkl3D IO
()
getDataFrameHkl3D d = do
  n <- lift $ hkl_h5_len (h5mu d)
  forM_ [0..n] $ \i -> yield $ getDataFrameHkl3D' d i
but when I compile it I get this error message
src/hkl3d.hs:274:3:
    Couldn't match type `IO DataFrameHkl3D' with `DataFrameHkl3D'
    Expected type: Proxy X () () DataFrameHkl3D IO ()
      Actual type: Proxy X () () (IO DataFrameHkl3D) IO ()
    In a stmt of a 'do' block:
      forM_ [0 .. n] $ \ i -> yield $ getDataFrameHkl3D' d i
    In the expression:
      do { n <- lift $ hkl_h5_len (h5mu d);
           forM_ [0 .. n] $ \ i -> yield $ getDataFrameHkl3D' d i }
    In an equation for `getDataFrameHkl3D':
        getDataFrameHkl3D d
          = do { n <- lift $ hkl_h5_len (h5mu d);
                 forM_ [0 .. n] $ \ i -> yield $ getDataFrameHkl3D' d i }
I do not know how to fix my code...
Frederic
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Re: [Haskell-beginners] pipe and IO

David McBride