Free-before-allocate in Java

[Shevek]

Given the following code:

byte[] data = ...;

{
data = null;
data = new byte[N];
}

Is the compiler allowed to discard the assignment of null, because it's
"dead" in language terms? My argument is that it isn't dead because it
allows the garbage collector to respond to pressure within the 'new' and
reuse the space, but in language terms, this presumably isn't defined,
and it would seem to be legal for the first assignment to be removed.

Thank you.

S.

[Jean-Philippe BEMPEL]

Hi,

JIT Compiler is allowed to discard the null assisgnment because there are 2 consecutives writes into the same location, so only the last one can be kept without changing the meaning of the program.

From the GC point of view as soon as the reference is no longer pointing to the allocate object, it is considered as dead, and may be reclaimed in the next cycle.

But depending on if data is a field or local variable will determine if GC can reclaim it sooner.

In the case of a local variable as soon as the variable is no more read, a GC can already collect the byte[] before assigning local variable to null (with the help of liveness analysis). 

For the field case, the byte[] will reclaimable once the assignment is done, so maybe after the allocation of the second byte[]

Regards

--
You received this message because you are subscribed to the Google Groups "mechanical-sympathy" group.
To unsubscribe from this group and stop receiving emails from it, send an email to mechanical-symp...@googlegroups.com.
For more options, visit https://groups.google.com/d/optout.

[Margaret Figura]

Hi,

I see why the compiler should be able to remove the first assignment, but (just for fun) I'm having trouble getting it to actually do it. With my -Xmx setting, I should see an OutOfMemoryError as soon as the first assignment is optimized away because the heap is large enough for only 1 copy of the new byte[] in the second assignment.

So, I'm wondering if this can actually happen. And if not, is it because the JIT just doesn't detect this possible optimization, or because it's smart enough to realize the trick I'm trying to play on it.

-Meg

To unsubscribe from this group and stop receiving emails from it, send an email to mechanical-sympathy+unsub...@googlegroups.com.

[Science Student]

Technically the 3rd line is not even needed for compiler to optimize away the 1st line. Thoughts?

To unsubscribe from this group and stop receiving emails from it, send an email to mechanical-symp...@googlegroups.com.

For more options, visit https://groups.google.com/d/optout.

--

You received this message because you are subscribed to the Google Groups "mechanical-sympathy" group.

To unsubscribe from this group and stop receiving emails from it, send an email to mechanical-symp...@googlegroups.com.

[Gil Tene]

This one is "tricky" in a fun way.

From a language semantics point of view, the compiler is probably allowed to perform the optimization suggested (removing the null assignment to a non-volatile field, knowing that it will soon be overwritten), which does raise an interesting point about the language semantics and OOM conditions.

However, from a practical implementation point of view, it is "hard" [my shorthand for "it may be impossible, but I can't prove it, or don't want to bother trying"] to perform this optimization in a way that would eliminate the null assignment from a GC perspective, because there is an object allocation between the two assignments. The difficulty comes from the fact that even in optimized code (in current JVM implementations), all allocation sites retain the ability to take a safepoint. Here is the logic:

- The allocation attempt *may* need to wait for a GC to complete (e.g. if a GC is needed in order to produce the empty memory that the allocation will use).

- A GC can't be guaranteed to complete (in all current practical JVM implementations) without transitioning all threads (at the very least temporarily and individually, if not simultaneously and globally) to a safepoint.

- Since the thread must be able to "come to a safepoint" at the allocation site (which sits between the first and second assignments), and since safepoints can end up being used for things other than GC (such as deoptimization, breakpoints, etc.), the JVM state at the safepoint must be completely reconstructible.

- If a safepoint is taken at the allocation site. the state of the JVM at that safepoint would include the memory outcome of the first assignment and *not* include the outcome second assignment.

- Therefore, *IF* the safepoint is taken, the null assignment must occur before it is taken.

Technically, this can either defeat the optimization altogether (which is what it will do for most JITs), or, it is possible to keep the optimization in the fast path (allocation doesn't take a safepoint). If a JIT is able to push code into the path between the poll to determine if a safepoint is needed and actually reaching the safepoint, the null assignment code could be moved around such that it occurs only if a safepoint is actually taken, and is skipped if a safepoint is not taken.

Either way (regardless of whether the optimization is defeated, or the null is moved to happen only in the safepoint-taking path), the null assignment would occur before GC is ever forced at the allocation site. This will explain why you won't see an OOM on that allocation on current JVMs, even with JIT-optimized code, even if the heap is only large enough to accommodate one copy of the byte[].