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Can non-overlapping spinlocks deadlock in C++?

Thursday, 15 April 2021

There has been discussion on Twitter recently about whether or not the C++ memory model allows spinlocks to deadlock if they just use memory_order_acquire in lock and memory_order_release in unlock, due to compiler optimizations. The case in question is where a thread locks one mutex, unlocks it, and locks a second: can the compiler reorder the second lock above the first unlock? If it does, and another thread does the same in the reverse order, with the same optimization, then sequential locks could deadlock.

Here is the code in question, with all the lock/unlock code inlined.

std::atomic<bool> mutex1{false};
std::atomic<bool> mutex2{false};

int x=0;
int y=0;

void thread1(){
  while(mutex1.exchange(true,std::memory_order_acquire)){}  // #1
  x=1;
  mutex1.store(false,std::memory_order_release); // #2

  while(mutex2.exchange(true,std::memory_order_acquire)){} // #3
  y=1;
  mutex2.store(false,std::memory_order_release); // #4
}

void thread2(){
  while(mutex2.exchange(true,std::memory_order_acquire)){} // #5
  x=2;
  mutex2.store(false,std::memory_order_release); // #6

  while(mutex1.exchange(true,std::memory_order_acquire)){} // #7
  y=2;
  mutex1.store(false,std::memory_order_release); // #8
}

For there to even be the possibility of deadlock, thread1 must successfully execute line #1 before thread2 successfully executes line #7, and thread2 must successfully execute line #5 before thread1 successfully executes line #3. Because these are RMW operations, the threads must agree on the ordering.

The modification order of mutex1 must thus be #1(success), #2, #7(success), #8. Similarly, the modification order of mutex2 must be #5(success), #6, #3(success), #4.

All threads must agree on these modification orders. https://eel.is/c++draft/intro.multithread#intro.races-4

From the point of view of thread1, everything must run in program order: compilers can only optimize things as long as they run "as if" in program order.

The store to mutex1 at #2 is guaranteed to be visible to thread2 in "a finite period of time". https://eel.is/c++draft/intro.multithread#intro.progress-18

Consequently, thread2 must eventually see that store at line #7, even if it executes line #7 a large number of times first.

Therefore, the compiler cannot move line #3 completely above line #2, since doing so would not guarantee the visibility of #2 to other threads in a finite period of time. It can move an arbitrary number of executions of line #3 above line #2 (all of which will see that mutex2 is still true), but not all the executions of line #3.

Given that thread2 eventually sees the store from #2 at line #7, the exchange at line #7 will eventually succeed, and thread2 will eventually complete.

Likewise, the store at #6 must become visible to thread1 in a finite period of time. Therefore the exchange at line #3 will eventually see the value stored by

6, the exchange will succeed, and thread1 will complete, and the compiler is

not allowed to move all the executions of line #7 above #6.

No amount of compiler optimization is allowed to break this, so no: spinlocks cannot deadlock if they don't overlap.

Posted by Anthony Williams
[/ cplusplus /] permanent link
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