A while ago, I wrote a post describing how unsafe FFI calls could block your entire system, and gave the following example of this behavior:

/* cbit.c */ #include <stdio.h> int bottom(int a) { while (1) {printf("%d

", a);sleep(1);} return a; }

/* cbit.h */ int bottom(int a);

/* UnsafeFFITest.hs */ {-# LANGUAGE ForeignFunctionInterface #-} import Foreign.C import Control.Concurrent main = do forkIO $ do safeBottom 1 return () yield print "Pass (expected)" forkIO $ do unsafeBottom 2 return () yield print "Pass (not expected)" foreign import ccall "cbit.h bottom" safeBottom :: CInt -> IO CInt foreign import ccall unsafe "cbit.h bottom" unsafeBottom :: CInt -> IO CInt

In the post, I explained that the reason this occurs is that unsafe FFI calls are not preemptible, so when unsafeBottom loops forever, the Haskell thread can't proceed.

This explanation would make perfect sense except for one problem: the code also hangs even when you run with the multi-threaded runtime system, with multiple operating system threads. David Barbour wrote in wondering if my claim that unsafe calls blocked the entire system was out of date. But the code example definitely does hang on versions of GHC as recent as 7.8.3. Based on the title of this post, can you guess the reason? If you think you know, what do these variants of the program do?

Change main = to main = runInUnboundThread Change the second forkIO to forkOn 2 Add a yield before unsafeBottom , and another yield before print "Pass (not expected)"

The reason why the code blocks, or, more specifically, why the main thread blocks, is because the unsafe FFI call is unpreemptibly running on the operating system thread which the main thread is bound to. Recall, by default, the main thread runs in a bound operating system thread. This means that there is a specific operating system thread which must be used to run code in main. If that thread is blocked by an FFI call, the main thread cannot run, even if there are other worker threads available.

We can thus explain the variants:

main is run in an unbound thread, no blocking occurs, and thus the second print runs. By default, a forked thread is run on the same capability as the thread that spawned it (this is good, because it means no synchronization is necessary) so forcing the bad FFI call to run on a different worker prevents it from blocking main. Alternately, if a thread yields, it might get rescheduled on a different worker thread, which also prevents main from getting blocked.

So, perhaps the real moral of the story is this: be careful about unsafe FFI calls if you have bound threads. And note: every Haskell program has a bound thread: main!