问题
Building with VS2013, specifying time_point::max()
to a condition variable's wait_until
results in an immediate timeout.
This seems unintuitive - I would naively expect time_point::max()
to wait indefinitely (or at least a very long time). Can anyone confirm if this is documented, expected behaviour or something specific to MSVC?
Sample program below; note replacing time_point::max()
with now + std::chrono::hours(1)
gives the expected behaviour (wait_for
exits once cv is notified, with no timeout)
#include <condition_variable>
#include <mutex>
#include <chrono>
#include <future>
#include <functional>
void fire_cv( std::mutex *mx, std::condition_variable *cv )
{
std::unique_lock<std::mutex> lock(*mx);
printf("firing cv\n");
cv->notify_one();
}
int main(int argc, char *argv[])
{
std::chrono::steady_clock::time_point now = std::chrono::steady_clock::now();
std::condition_variable test_cv;
std::mutex test_mutex;
std::future<void> s;
{
std::unique_lock<std::mutex> lock(test_mutex);
s = std::async(std::launch::async, std::bind(fire_cv, &test_mutex, &test_cv));
printf("blocking on cv\n");
std::cv_status result = test_cv.wait_until( lock, std::chrono::steady_clock::time_point::max() );
//std::cv_status result = test_cv.wait_until( lock, now + std::chrono::hours(1) ); // <--- this works as expected!
printf("%s\n", (result==std::cv_status::timeout) ? "timeout" : "no timeout");
}
s.wait();
return 0;
}
回答1:
I debugged MSCV 2015's implementation, and wait_until
calls wait_for
internally, which is implemented like this:
template<class _Rep,
class _Period>
_Cv_status wait_for(
unique_lock<mutex>& _Lck,
const chrono::duration<_Rep, _Period>& _Rel_time)
{ // wait for duration
_STDEXT threads::xtime _Tgt = _To_xtime(_Rel_time); // Bug!
return (wait_until(_Lck, &_Tgt));
}
The bug here is that _To_xtime
overflows, which results in undefined behavior, and the result is a negative time_point
:
template<class _Rep,
class _Period> inline
xtime _To_xtime(const chrono::duration<_Rep, _Period>& _Rel_time)
{ // convert duration to xtime
xtime _Xt;
if (_Rel_time <= chrono::duration<_Rep, _Period>::zero())
{ // negative or zero relative time, return zero
_Xt.sec = 0;
_Xt.nsec = 0;
}
else
{ // positive relative time, convert
chrono::nanoseconds _T0 =
chrono::system_clock::now().time_since_epoch();
_T0 += chrono::duration_cast<chrono::nanoseconds>(_Rel_time); //Overflow!
_Xt.sec = chrono::duration_cast<chrono::seconds>(_T0).count();
_T0 -= chrono::seconds(_Xt.sec);
_Xt.nsec = (long)_T0.count();
}
return (_Xt);
}
std::chrono::nanoseconds
by default stores its value in a long long
, and so after its definition, _T0
has a value of 1'471'618'263'082'939'000
(this changes obviously). Adding _Rel_time
(9'223'244'955'544'505'510
) results definitely in signed overflow.
We have already passed every negative time_point
possible, so a timeout happens.
来源:https://stackoverflow.com/questions/39041450/stdcondition-variable-wait-until-surprising-behaviour