[Haskell-cafe] Timing out a pure evaluation of an expression I did not write myself

Daniel Díaz Casanueva dhelta.diaz at gmail.com
Sun Nov 18 00:51:58 UTC 2018

Hello Ryan.

Try evaluating the expression to normal form instead of weak head normal
form in your expression. So:

>>> timeout 1 $ evaluate $ force $ let x = 0 : x in last x

The function `force` comes from the deepseq package. You can read the docs

I hope that helps.

Best regards,

Am So., 18. Nov. 2018 um 00:22 Uhr schrieb Ryan Reich <ryan.reich at gmail.com

> I want to time out a pure computation.  My experience, and that described
> in various previous questions here and elsewhere (the best of which is
> https://mail.haskell.org/pipermail/haskell-cafe/2011-February/088820.html),
> is that this doesn't always work: for instance,
> >>> timeout 1 $ evaluate $ let x = 0 : x in last x
> does not time out because, apparently, the fact that the expression
> evaluates in constant space (i.e. never allocates) means that it never
> yields to the timeout monitor thread that would kill it.
> The solution that is described in the other iterations is to embed
> checkpoints in the expression that do allocate, giving the RTS a chance to
> switch contexts.  However, in my application, the expression is /arbitrary/
> and I do not have the freedom to inject alterations into it.  (Don't argue
> this point, please.  The expression is arbitrary.)
> How can I time out a tight loop like the above?  Clearly, it can be done,
> because I can, say, alt-tab over to another terminal and kill the process,
> which exploits the operating system's more aggressively pre-emptive
> scheduling.  Is there a solution using bound threads, say 'forkOS' instead
> of 'forkIO' in the implementation of 'timeout'?  Unix signals?  Some
> FFI-based workaround?  Etc.  Keep in mind that notwithstanding that
> comment, I don't actually want to kill the whole process, but just the one
> evaluation.
> Thanks in advance,
> Ryan Reich
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