megamorphic lambda prevention

[Rémi Forax]

On 06/21/2012 04:38 PM, Matt Fowles wrote:
> All~
>
> Couldn't the VM detect hot mega-morphic methods and have them bestow 
> some of their hotness upon the callers.  Making their callers more 
> likely to inline them.  Then you have a different call site for the 
> each which is likely to be monomorphic.
>
> Matt

yes, that the idea when i said the the VM can go backward.
Note that it works only for callsites that are megamorphic because of a 
value
that comes from the parameters of the high order method.

There is also a special case for OSR because in that case you may have 
the stack
so it can be easier to go backward.

Rémi

>
> On Thu, Jun 21, 2012 at 10:30 AM, Rémi Forax <forax at univ-mlv.fr 
> <mailto:forax at univ-mlv.fr>> wrote:
>
>     On 06/21/2012 04:12 PM, Jochen Theodorou wrote:
>     > Am 21.06.2012 16 <tel:21.06.2012%2016>:00, schrieb Rémi Forax:
>     >> On 06/21/2012 03:52 PM, Jochen Theodorou wrote:
>     >>> Am 21.06.2012 13 <tel:21.06.2012%2013>:57, schrieb Krystal Mok:
>     > [...]
>     >>>> I wonder how well the new interpreter design in Graal would
>     handle this
>     >>>> kind of case, since it's supposed to have picked the good
>     parts from
>     >>>> trace-based compilation, but without actually having to do
>     tracing.
>     >>>> Can't wait to see more details of it at this year's JVM
>     Language Summit.
>     >>> If I understood Thomas right, then Graal doesn't have that
>     problem.
>     >>> Afaik hotspot doesn't have to have that problem too. It just was a
>     >>> design decision not to make call site caching per execution
>     path. Afaik
>     >>> chaning that now would take quite some effort, but is not
>     impossible.
>     >> You can create the execution path backward, when it's needed.
>
>     I was thinking that the VM can create the path backward.
>
>     > you can in PHP.reboot, but in Groovy I don't have access to the full
>     > execution path. Inside "each" I don't know what did call me and
>     at the
>     > point of where I call each, I don't see what each will actually do.
>
>     yes, but you can detect that 'each' is a special method because
>     it takes a closure (Closure) as parameter and try to do some magic.
>
>     >
>     > bye blackdrag
>     >
>
>     Rémi
>
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