Unmask readMVar in readChan

When `readMVar` was implemented using `takeMVar` and `putMVar`,
we needed to use `modifyMVarMasked` in `readChan` just in case
the `readMVar` was interrupted between taking and putting. Now
that `readMVar` uses an atomic primop, this is impossible, so we can
safely unmask `readMVar`.

Reviewers: hvr, bgamari, simonmar

Reviewed By: simonmar

Subscribers: rwbarton, thomie, carter

Differential Revision: https://phabricator.haskell.org/D4738

1 files changed, 1 insertions, 10 deletions

diff --git a/libraries/base/Control/Concurrent/Chan.hs b/libraries/base/Control/Concurrent/Chan.hs
index d752a898f0..d0fed4a405 100644
--- a/libraries/base/Control/Concurrent/Chan.hs
+++ b/libraries/base/Control/Concurrent/Chan.hs
@@ -106,21 +106,12 @@ writeChan (Chan _ writeVar) val = do
 -- thread holds a reference to the channel.
 readChan :: Chan a -> IO a
 readChan (Chan readVar _) = do
-  modifyMVarMasked readVar $ \read_end -> do -- Note [modifyMVarMasked]
+  modifyMVar readVar $ \read_end -> do
     (ChItem val new_read_end) <- readMVar read_end
         -- Use readMVar here, not takeMVar,
         -- else dupChan doesn't work
     return (new_read_end, val)
 
--- Note [modifyMVarMasked]
--- This prevents a theoretical deadlock if an asynchronous exception
--- happens during the readMVar while the MVar is empty.  In that case
--- the read_end MVar will be left empty, and subsequent readers will
--- deadlock.  Using modifyMVarMasked prevents this.  The deadlock can
--- be reproduced, but only by expanding readMVar and inserting an
--- artificial yield between its takeMVar and putMVar operations.
-
-
 -- |Duplicate a 'Chan': the duplicate channel begins empty, but data written to
 -- either channel from then on will be available from both.  Hence this creates
 -- a kind of broadcast channel, where data written by anyone is seen by