I'm quite certain that code is correct. There's no data race here since the write and the non-atomic read are both inside the mutex.
The memory model rules about mutexes formalize how they serialize the code inside them. Because the unlock happens-before the next lock, and because sequential code in a single goroutine has the happens-before order that the source code expresses, this read: if o.done == 0 and this write: atomic.StoreUint32(&o.done, 1) are not concurrent. (This is just a slightly more precise way of stating the natural, intuitive behavior of a mutex.) Therefore, that read *must* observe the previously written value. On Thu, Dec 28, 2017 at 7:11 PM, Dave Cheney <d...@cheney.net> wrote: > I think you are right, all access to that field should go via atomic > load/store operations. I don’t think it’s safe to mix those operations wth > normal loads and stores even within the same goroutine. > > -- > You received this message because you are subscribed to the Google Groups > "golang-nuts" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to golang-nuts+unsubscr...@googlegroups.com. > For more options, visit https://groups.google.com/d/optout. -- You received this message because you are subscribed to the Google Groups "golang-nuts" group. To unsubscribe from this group and stop receiving emails from it, send an email to golang-nuts+unsubscr...@googlegroups.com. For more options, visit https://groups.google.com/d/optout.
