Issue #20314 has been updated by jeremyevans0 (Jeremy Evans). Solution 2 makes the most sense to me. If inside a `Timeout.timeout` block, you are swallowing the exception that you provided in the `Timeout.timeout` method call, that to me indicates you do not want to handle it as a timeout. ---------------------------------------- Bug #20314: Simultaneous Timeout expires may raise an exception after the block https://bugs.ruby-lang.org/issues/20314#change-107127 * Author: mame (Yusuke Endoh) * Status: Open * Backport: 3.0: UNKNOWN, 3.1: UNKNOWN, 3.2: UNKNOWN, 3.3: UNKNOWN ---------------------------------------- Launchable reports `TestTimeout#test_nested_timeout` as a flaky test, and I reproduced it as follows. ```ruby require "timeout" class A < Exception end class B < Exception end begin Timeout.timeout(0.1, A) do Timeout.timeout(0.1, B) do nil while true end end rescue A, B p $! #=> #<A: execution expired> # Exception B is raised after the above call returns #=> test.rb:16:in `p': execution expired (B) p :end # not reach end ``` This is because the timer thread performs two consecutive `Thread#raise` to the target thread. I have discussed this with @ko1 and have come up with three solutions. ### Solution 1 When multiple nested Timeouts expire simultaneously, raise an exception for the outer-most Timeout and let the inner Timeouts expire without throwing an exception. In the above example, it would only raise A. The problem with this approach is that if you are rescuing A in the inner block, it may never ends: ```ruby Timeout.timeout(0.1, A) do Timeout.timeout(0.1, B) do begin sleep rescue A sleep # The exception A is caught. The inner Timeout is already expired, so the code (may) never end. end end end ``` Note that, if A and B did not occur at the same time, it would raise B. This is a race condition. ### Solution 2 When multiple nested Timeouts expire simultaneously, raise an exception for the inner-most Timeout and let the outer Timeouts wait until the inner-most Timeout returns. In the above example, it would raise either A or B, not both. The problem with this approach is that if you are rescuing B in the inner block, it never ends: ```ruby Timeout.timeout(0.1, A) do Timeout.timeout(0.1, B) do begin sleep rescue B sleep # The outer Timeout waits for the inner timeout, and the inner Timeout never return. So this code never ends. end end end ``` ### Solution 3 Make thread interrupt queue one length. If the target thread has already been `Thread#raise(A)`, the new `Thread#raise(B)` blocks until the target thread processes A. Since there will be no more simultaneous Thread#raise, there will be no more exceptions after the end of the block. The timeout timer thread should be changed in consideration that `Thread#raise` may block. -- https://bugs.ruby-lang.org/