But, f1 and f2 are already declared as val. So, there should not be problem in
accessing them.
Further, both the async statements just make it explicit what is already
implicit.
async { f1 = fib(n-1); } implicitly means that async activity is created here,
where the root activity resides, and the enclosed statement is also executed
here.
The statements at (here) async { f1 = fib(n-1); } and async at (here) {fi =
fib(n-1); } must also start new activity here and execute the enclosed
statement in the current place, i.e., here.
So, shouldn't all these three forms be correct?
Jeeva P.
Re: [X10-users] async at (here) and at (here) async
Yoav Zibin
Fri, 06 May 2011 10:22:37 -0700
This is a bug.
In the upcoming 2.2 release both attempts will fail with the error:
Local variable is accessed at a different place, and therefore it must be
initialized and declared final.
So you cannot access a local variable in a different place unless it is
initialized and "val".
In 2.3 there will be a new construct called "athome" that will allow you to
initialize a local at a remote place by going back to the original place.
On Thu, May 5, 2011 at 11:45 PM, Jeeva Paudel <je...@ualberta.ca> wrote:
> Hi all,
>
> This is in reference to the Fibonacci example available at
> http://dist.codehaus.org/x10/applications/samples/Fib.x10
>
> Although async at != at async always, but shouldn't the following two
> versions of async produce similar effect?
>
> at (here) async { f1 = fib(n-1); } must start a new activity located
> here and executing S
>
> async at (here) {fi = fib(n-1); } must also start a new activity
> here and execute S here as well.
>
> Aren't both the same and should compile correctly when the program
> compiles correctly with async { f1 = fib(n-1)} ? However, the compiler
> reports this error : "f1" may not have been initialized
>
>
> Thanks,
> Jeeva P.
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