Why does the JVM require warmup?

Question

I understand that in the Java virtual machine (JVM), warmup is potentially required as Java loads classes using a lazy loading process and as such you want to ensure that th

Answer 1

Warming refers to having a piece of code run enough times that the JVM stops interpreting and compiles to native (at least for the first time). Generally that's something you don't want to do. The reason is that the JVM gathers statistics about the code in question that it uses during code generation (akin to profile guided optimizations). So if a code chunk in question is "warmed" with fake data which has different properties than the real data you could well be hurting performance.

EDIT: Since the JVM cannot perform whole-program static analysis (it can't know what code is going to be loaded by the application) it can instead make some guesses about types from the statistics it has gathered. As an example when calling a virtual function (in C++ speak) at an exact calling location and it determines that all types have the same implementation, then the call is promoted to direct call (or even inlined). If later that assumption if proven to be wrong, then the old code must be "uncompiled" to behave properly. AFAIK HotSpot classifies call-sites as monomorphic (single implementation), bi-morphic (exactly two..transformed into if (imp1-type) {imp1} else {imp2} ) and full polymorphic..virtual dispatch.

And there's another case in which recompiling occurs..when you have tiered-compilation. The first tier will spend less time trying to produce good code and if the method is "hot-enough" then the more expensive compile-time code generator kicks in.