Re: volatile long

> each is composed of a read followed by a write.

I think this is false.  Had the statements been something like:

if (count++) ...

Then it would be a read followed by a write.  But just count++ only
involves a write.

Hanson

On Thu, 17 Feb 2005 01:10:43 -0800, Joe Bowbeer <jozart@xxxxxxxxx> wrote:
> Even though though ++ and -- are each represented by a single bytecode
> (iinc), I don't think these are atomic operations in terms of the JMM.  That
> is, each is composed of a read followed by a write.  In which case, even an
> "int" count can creep very negative or very positive over time, depending on
> the interleaving of threads.  Use AtomicInteger to prevent counter creep.
> 
> Volatile protects long values from word shearing, but it doesn't turn ++
> or -- into atomic operations either.
> 
> 
> ----- Original Message -----
> From: "Hanson Char" <hanson.char@xxxxxxxxx>
> To: <concurrency-interest@xxxxxxxxxxxxxxxxxxxx>
> Sent: Thursday, February 17, 2005 12:49 AM
> Subject: [concurrency-interest] volatile long
> 
> A question on the use of volatile in a concurrent environment.
> Consider the example:
> 
> public class  Foo {
>     private volatile int count;
> 
>      public void run() {
>         count++;
>         // ... do some other operations
>         count--;
>     }
>      public int getCount() { return count; }
> }
> 
> My understanding is:
> 
> 1) The value returned by getCount() will never be less than zero; and
> 2) count will never be in a corrupted state.
> 
> Does the above 2 statements still hold if count is type long instead of int
> ?
> 
> Hanson
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