I think the reason is that no memory allocation occurs in the loop you are trying to interrupt. Thus, the RTS can not stop the Haskell thread to terminate the program after catching the signal. You can only stop a Haskell thread if you are at a safe-point (i.e. memory allocation). If you call a function like 'last', which is already compiled, you get this behavior (you can kill it if you define last in ghci itself, because then it is interpreted instead of using the compiled version). There is a flag (-fno-omit-yields) to compile a module with additional safe-points if there is no memory allocation going on in a loop, but base is build without it, so you get the behavior you observe. On 05/18/2017 09:09 AM, Jeremy Henty wrote:
If I put ghci into a loop by entering "last $ repeat ()" then the process is impossible to interrupt or kill except with "kill -KILL". (Although it can be stopped with Ctrl-Z.) "kill -TERM" does not work. The same is true if I run a runhaskell script containing "main = putStrLn $ show $ last $ repeat $ ()".
Is there a reason for ghci being so unkillable? It is generally considered a bad idea for a program to ignore the TERM signal.
(Details: Debian Linux Jessie 8.8, GHC 8.0.2 (built from source).)
Regards,
Jeremy Henty _______________________________________________ Haskell-Cafe mailing list To (un)subscribe, modify options or view archives go to: http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe Only members subscribed via the mailman list are allowed to post.