是std :: atomic <T> 当std :: atomic时中断安全 <T> :: is_always_lock_free是假的？

Question

In an embedded (ARM) environment with no OS, if I use interrupts, then is there potential for deadlock using std::atomic<T> ? If so, how?

In general, any moment, control can be interrupted to handle an interrupt. In particular, if one were to naively have a mutex and wanted to use it to do a "safe" to a variable, one might lock it, write, and unlock and then elsewhere lock, read, and unlock. But if the read is in an interrupt, you could lock, interrupt, lock => deadlock.

In particular, I have a std::atomic<int> for which is_always_lock_free is false . Should I worry about the deadlock case? When I look at the generated assembly, writing 42 looks like:

bl __sync_synchronize
mov r3, #42
str r3, [sp, #4]
bl __sync_synchronize

which doesn't appear to be locking. The asm for reading the value is similar. Is the (possible) lock for the fancier operations like exchange ?

Answer 1

__sync_synchronize is just a builtin for a full memory barrier. There is no locking involved, so no potential for deadlock as there would be with a mutex and interrupt handler.

What ARM core are you using? On an ARM Cortex-A7 the following prints true for both.

#include <iostream>
#include <atomic>

int main()
{
   std::atomic<int> x;
   std::cout << std::boolalpha << x.is_lock_free() << std::endl;
   std::cout << std::atomic<int>::is_always_lock_free << std::endl;
}

I would expect std::atomic<int> to be implemented without locks most if not all on ARM, and certainly from the assembly you provided it does not appear to be using a lock.