On Wed, Feb 12, 2014 at 10:19:07AM +0100, Peter Zijlstra wrote: > > I don't know the specifics of your example, but from how I understand > > it, I don't see a problem if the compiler can prove that the store will > > always happen. > > > > To be more specific, if the compiler can prove that the store will > > happen anyway, and the region of code can be assumed to always run > > atomically (e.g., there's no loop or such in there), then it is known > > that we have one atomic region of code that will always perform the > > store, so we might as well do the stuff in the region in some order. > > > > Now, if any of the memory accesses are atomic, then the whole region of > > code containing those accesses is often not atomic because other threads > > might observe intermediate results in a data-race-free way. > > > > (I know that this isn't a very precise formulation, but I hope it brings > > my line of reasoning across.) > > So given something like: > > if (x) > y = 3; > > assuming both x and y are atomic (so don't gimme crap for now knowing > the C11 atomic incantations); and you can prove x is always true; you > don't see a problem with not emitting the conditional? You need volatile semantics to force the compiler to ignore any proofs it might otherwise attempt to construct. Hence all the ACCESS_ONCE() calls in my email to Torvald. (Hopefully I translated your example reasonably.) Thanx, Paul > Avoiding the conditional changes the result; see that control dependency > email from earlier. In the above example the load of X and the store to > Y are strictly ordered, due to control dependencies. Not emitting the > condition and maybe not even emitting the load completely wrecks this. > > Its therefore an invalid optimization to take out the conditional or > speculate the store, since it takes out the dependency. > -- To unsubscribe from this list: send the line "unsubscribe linux-arch" in the body of a message to majordomo@xxxxxxxxxxxxxxx More majordomo info at http://vger.kernel.org/majordomo-info.html
