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Fix infinite wait bug in test_pthread_proxying_cpp.cpp #18358

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Merged
tlively merged 1 commit into from
Dec 13, 2022
Merged

Fix infinite wait bug in test_pthread_proxying_cpp.cpp #18358

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20 changes: 16 additions & 4 deletions test/pthread/test_pthread_proxying_cpp.cpp
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Original file line number Diff line number Diff line change
Expand Up @@ -56,7 +56,10 @@ void test_proxy_async() {
// Proxy to looper.
{
queue.proxyAsync(looper.native_handle(), [&]() {
i = 2;
{
std::unique_lock<std::mutex> lock(mutex);
i = 2;
}
executor = std::this_thread::get_id();
cond.notify_one();
});
Expand All @@ -68,7 +71,10 @@ void test_proxy_async() {
// Proxy to returner.
{
queue.proxyAsync(returner.native_handle(), [&]() {
i = 3;
{
std::unique_lock<std::mutex> lock(mutex);
i = 3;
}

@sbc100 sbc100 Dec 13, 2022

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Despite reading you description I've having trouble seeing how putting a lock around this one assignment fixes the issue. Would making i and atomic also fix the issue?

@sbc100 sbc100 Dec 13, 2022

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After reading https://stackoverflow.com/questions/17101922/do-i-have-to-acquire-lock-before-calling-condition-variable-notify-one I think i understand.. I had never known this about cond vars..

@tlively tlively Dec 13, 2022

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No, the lock is important because of the condition variable. The condition variable needs to atomically check the condition and block if the condition is not met. The mechanism for combining those into an atomic action is the lock, so if the condition can change without the lock being held, the condition variable doesn't work as intended.

executor = std::this_thread::get_id();
cond.notify_one();
});
Expand Down Expand Up @@ -164,7 +170,10 @@ void test_proxy_async_with_callback(void) {
queue.proxyAsyncWithCallback(
looper.native_handle(),
[&]() {
i = 2;
{
std::unique_lock<std::mutex> lock(mutex);
i = 2;
}
executor = std::this_thread::get_id();
cond.notify_one();
},
Expand All @@ -184,7 +193,10 @@ void test_proxy_async_with_callback(void) {
queue.proxyAsyncWithCallback(
returner.native_handle(),
[&]() {
i = 3;
{
std::unique_lock<std::mutex> lock(mutex);
i = 3;
}
executor = std::this_thread::get_id();
cond.notify_one();

@sbc100 sbc100 Dec 13, 2022

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You don't need to hold the lock when calling notify too?

@tlively tlively Dec 13, 2022

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No, it's actually a pessimization to hold the lock while calling notify because then the other thread wakes up just to immediately block on acquiring the lock.

@sbc100 sbc100 Dec 13, 2022

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TIL!

},
Expand Down