[ruby-core:117003] [Ruby master Bug#20314] Simultaneous Timeout expires may raise an exception after the block
Issue #20314 has been reported by mame (Yusuke Endoh). ---------------------------------------- Bug #20314: Simultaneous Timeout expires may raise an exception after the block https://bugs.ruby-lang.org/issues/20314 * 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/
Issue #20314 has been updated by Eregon (Benoit Daloze). I'm not sure how Solution 3 would work. `Thread#raise` would block until what? Until the exception started to be raised/thrown on that thread? I think that would not fix that snippet. It does not seem reasonable to wait until the exception has been rescued (or escapes the thread) because that could run arbitrary code via `ensure` which could take a long time (and block the caller of `Thread#raise` for a long time). I wonder if we should always use `Timeout::ExitException` to "unwind" until we exit the corresponding block given to `Timeout.timeout`. Unsure if that would help for this issue though. Maybe there are other solutions too? Between solutions 1 and 2, 2 seems better because it seems clearly broken code to `rescue B; sleep; end` in `Timeout.timeout(0.1, B)`. In any case `sleep` in `ensure`/`rescue` is broken code as it can already hang (e.g. with a single `Timeout.timeout(0.1, SomeKlass)` around it, or without using Timeout). ---------------------------------------- Bug #20314: Simultaneous Timeout expires may raise an exception after the block https://bugs.ruby-lang.org/issues/20314#change-107071 * 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/
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/
Issue #20314 has been updated by mame (Yusuke Endoh). Indeed Solution 3 seems to be wrong. I think I must have misunderstood what ko1 was saying. Actually, I don't like neither Solutions 1 and 2. In the code example, I used exceptions A and B just for clarity. In most real-world cases, both are `Timeout::Error`. Swallowing `Timeout::Error` is a normal practice (to make `Timeout.timeout` end gracefully), so I don't think it is "broken". I think this is a design flaw of Timeout's API, but I don't know what we can do for that. I guess we will just have to go with Solution 2 for now? ---------------------------------------- Bug #20314: Simultaneous Timeout expires may raise an exception after the block https://bugs.ruby-lang.org/issues/20314#change-107133 * 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/
Issue #20314 has been updated by Eregon (Benoit Daloze). Ref: https://github.com/ruby/ruby/pull/10851 ---------------------------------------- Bug #20314: Simultaneous Timeout expires may raise an exception after the block https://bugs.ruby-lang.org/issues/20314#change-110168 * 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/
participants (3)
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Eregon (Benoit Daloze) -
jeremyevans0 (Jeremy Evans)
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mame (Yusuke Endoh)