On May 10, 2013, at 1:37 AM, Jochen Theodorou <blackd...@gmx.org> wrote:
> Am 10.05.2013 02:40, schrieb Christian Thalinger:
> [...]
>> That's because your method handle is not constant and so the compiler cannot
>> inline the call.
>
> And you tell me that in the first case the call was inlined?
No. I was referring to your comment about "hotspot should throw out the
invocation completely, but it does not". That's the case in both examples.
> That is
> unexpected. And if that is the case, then why does this:
>
>>>> MethodHandle mh = MethodHandles.lookup().findVirtual(String.class,
>>>> "toString", MethodType.methodType(String.class));
>>>> MethodType invokeType = MethodType.methodType(Object.class,
>>>> String.class);
>>>> int tmax = 1_000_000;
>>>> for (int k=0; k<20; k++) {
>>>> long t1 = System.nanoTime();
>>>> for (int i=0; i<tmax; i++) {
>>>> MethodHandle invoker = mh.asType(invokeType);
>>>> }
>>>> long t2 = System.nanoTime();
>>>> System.out.println((t2-t1)/tmax);
>>>> }
>
> take that long as well? Maybe it changed, but I thought only handles
> coming from a final field are constant enough for inlining. If I compare
> to 1.7.11 (or 1.7.6) then I get around 23 for the first, and 200-300 for
> the second and third test. That is about a factor 10-15 slowdown, that
> is what I would expect from the call not being inlined for example. But
> I always thought 1.7 will not inline in this case at all. On my 1.8 it
> was almost a factor 100. So unless the noninlined call on a method
> handle got extremely expensive compared to 1.7, there must be something
> else too.
I haven't looked but maybe the slowdown is because the asType call cannot be
inlined completely and so the compiler doesn't know it always returns the same
result.
-- Chris
>
> bye blackdrag
>
> --
> Jochen "blackdrag" Theodorou - Groovy Project Tech Lead
> blog: http://blackdragsview.blogspot.com/
> german groovy discussion newsgroup: de.comp.lang.misc
> For Groovy programming sources visit http://groovy-lang.org
>
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