Oliver Kullmann wrote:
Given two functions
bool f1(X x); bool f2(X x);
I want to compute the logical or (short-circuit!).
Now boost::thread did not support thread cancellation or termination in version 1_32, and it seems that development of boost::thread has been cancelled itself. To cancel a thread is essential to gain the speed-up possible by the parallel short-circuit evaluation, and thus boost::thread seems not to be usable.
Now how are you proceeding? In the FAQ's of boost::thread one can read
----
9. Why isn't thread cancellation or termination provided?
There's a valid need for thread termination, so at some point probably will include it, but only after we can find a truly safe (and portable) mechanism for this concept.
----
How do you cancel threads?
The explanation in the FAQ doesn't explain the problem in details. Here's a brief example: bool f1(X x) { X * y = new X; // if f1() is aborted at this point // *y won't be deallocated delete y; return false; }
Can one compute the short-circuit parallel or with your "futures" ?!
This is a common problem in the world of parallel computations. It's due to the nature of parallel exectution and it's independent from boost::thread, futures or any library. Basically, the solution is always the same: f1() polls a flag periodically, and if it is raised it exits. To cancel the thread you just raise the flag and the thread terminates itself: class A { Flag cancel_f1; bool f1(X x) { X * y = new X; for (int i = 1; i < 1000; i++) { // some computations, e.g. one iteration of the algorithm if (flag.is_raised()) { delete y; return false; } } return actual_result; } } ... A a; a.run_f1_in_a_new_thread(); a.cancel_f1.raise();
3) How is access to common storage handled with you approach?! (I would appreciate a short explanation using not too much jargon --- yet I have only written one multi-threaded program (and that to explore the boost::thread library).)
It isn't handled at all. You should use boost::thread synchronization primitives to handle it properly when you implement f1(). Andrey