Question

We have different JRE implementation from 2 vendors one is Jrockit from BEA and other is from Sun. In my project the document clearly states to use the Jrockit implementation from BEA. So i want to know the differences and upperhand of Jrockit over the standard sun JVM.

thanks Punith

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Solution

Firstly, BEA and Sun were both taken over by Oracle. So JRockit and HotSpot are now both Oracle products.

JRockit started out as a faster JVM than Hotspot for server-side code, but a lot of work has been done since then to make Hotspot faster, so it is not clear if that still applies. The other thing that I recall is that JRockit has (had) a different heap sizing strategy. In particular, you didn't need to provide a fixed upper bound for the heap. This could be a good thing or a bad thing, depending on your perspective.

From a purely functional stand-point, JRockit and HotSpot implementations of the same Java baseline should be virtually identical.

Here are some other resources (from a Google search):

OTHER TIPS

Technically they are both from Oracle now. ;)

Given most of the developers for the former Sun JVM are from the JRockit team, I suspect the difference will become smaller over time (i.e. there is only one development team for both products)

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