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