transparent lambda

[Zdenek Tronicek]

Did you have a look at the closures prototype? I think you describe the
way how non-local transfer is implemented.

Z.
-- 
Zdenek Tronicek
FIT CTU in Prague


Peter Levart napsal(a):
> On Monday 04 January 2010 11:07:45 Peter Levart wrote:
>> Here's another example:
>>
>> #int() one = null;
>> synchronized int sum(int n) {
>>   if (one == null) one = #() { return 1; };
>>   return n == 0 ? 0 : sum(n-1) + one();
>> }
>>
>
>> The compiler could generate the following code for the above example:
>>
>
> Ops, not entirely correct.
>
> So here's the first "refinement" that should make the generated code
> correct (changes to previous code are: try/catch wraps the whole method
> body not each lambda invocation):
>
> #int() one = null;
> synchronized int sum(int n) {
>   #int() $tmpLambda = null;
>   try {
>
>     if (one == null) {
>       $tmpLambda = new #int()() {
>         public int invoke() { throw new NonLocalReturnInt(this, 1); }
>       };
>       one = $tmpLambda;
>     }
>
>     return n == 0 ? 0 : sum(n-1) + one();
>
>   }
>   catch (NonLocalReturnInt $nlr) {
>     if ($tmlLambda == $nlr.lambda)
>       return $nlr.retval;
>     else
>       throw $nlr;
>   }
> }
>
> Each lambda expression that contains a non-local return statement and is
> specified in a method or constructor reserves a "$tmpLambda" slot on the
> stack which is filled-in with a reference to the lambda when the lambda
> expression is evaluated. The "NonLocalReturn" exception also contains a
> reference to the lambda that initiated the non-local transfer. The body of
> each method/constructor that contains lambda expressions with non-local
> returns is wrapped in a try/catch that catches the "NonLocalReturn" and
> checks it's "lambda" reference with all the "$tmpLambda" slots. If any of
> them matches, "NonLocalReturn" is translated into a return from the
> method/constructor else it is re-thrown.
>
> Regards, Peter
>