As a workaround, I am able to trick the child haskell process into thinking it's running in an interactive terminal by wrapping it with /usr/bin/script:

createProcess (proc "/usr/bin/script" ["-qc","./A","/dev/null"]) { std_out = CreatePipe }

I still think haskell is using screwy defaults for stdout buffering..

On Thursday, July 31, 2014 3:24:47 PM UTC-4, Chris Myzie wrote:

Hello,

I'm trying to write a wrapper process that will be able to read any child process' stdout.  The problem I'm running into is that unless I force the child's stdout to LineBuffering, it defaults to BlockBuffering.  Is BlockBuffering really the default in this case?  I don't want to have to modify all of the child processes that I want to use with this wrapper.

Below is a simple test case.  A.hs is the child process, and B.hs is the wrapper.  If I run B.hs, I will get no output unless I uncomment the line in A.hs.

Thanks,
Chris


------------------------------ A.hs ---------------------------
import Control.Concurrent
import System.IO

main :: IO ()
main = do
  -- hSetBuffering stdout LineBuffering
  putStrLn "test" >> threadDelay 1000000 >> main


------------------------------ B.hs ---------------------------
import Control.Monad
import System.IO
import System.Process

main :: IO ()
main = do
  (_,Just h,_,_) <- createProcess (proc "./A" []) { std_out = CreatePipe }
  hSetBuffering h LineBuffering
  forever $ hGetLine h >>= putStrLn