Re: JavaMemoryModel: PC language alert: catatonia action

From: Sylvia Else (
Date: Mon May 17 2004 - 22:57:38 EDT

Huge cc list trimmed because everyone seems to be on this list anyway.

At 02:39 PM 17/05/2004 -0400, Bill Pugh wrote:

>Essentially, it gives us three guarantees:
>* if there exists some thread that is able to make progress, some thread
>will make progress
>* compilers cannot, in general, hoist volatile reads out of potentially
>infinite loops
>* compilers cannot, in general, reorder synchronization actions and
>externally visible actions

I'm unclear as to how the catatonia action achieves the second of them, in

Taking the original example

v is volatile and initially v = 0
Thread 1:
while(v == 0);
println("thread 1 done");

Thread 2:
v = 1
println("thread 2 done");

My current understanding is that the v == 0 test cannot be hoist from the
loop because doing so would leave thread 1 in an infinite loop that does
nothing, but the program contains no catatonia action, so the resulting
program does not fit the model.

But consider

i not volatile.
Thread 1:
while(v == 0) {
println("thread 1 done");

What's to stop the hoist now? The resulting loop would still contain memory
operations, so no catatonia action is required, and there are no hb
relationships that would be violated that I can see.

Alternatively, could you explain how the existence in the model of
catatonia actions prohibits the volatile read hoist in the second case.


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