Using Java volatile keyword in non-thread scenario - java

I understand that the Java keyword volatile is used in multi-threading context; the main purpose is to read from the memory rather than from the cache or even if read from the cache, it would be updated first.
In the below example, there is no multi-threading concept. I want to understand if the variable i would be cached as a part of code optimization and hence read from cpu cache rather than memory? If yes, if the variable is declared as volatile, will it certainly be read from the memory?
I have run the program multiple times, by adding and also by deleting the volatile keyword; but, since there is no constant time for the for loop, I was unable to come to a conclusion if more time is consumed when the variable is declared as volatile.
All I want to see is that the time taken from CPU cache is actually less than when it is declared as volatile.
Is my understanding even right? If yes, how can I see the concept in working, with a good record of the times for both CPU cache reads and memory reads?
import java.time.Duration;
import java.time.Instant;
public class Test {
volatile static int i=0;
// static int i=0;
public static void main(String[] args) {
Instant start = Instant.now();
for (i=0; i<838_860_8; i++) { // 2 power 23; ~ 1 MB CPU Cache
System.out.println("i:" + i);
}
Instant end = Instant.now();
long timeElapsed = Duration.between(start, end).getSeconds();
System.out.println("timeElapsed: " + timeElapsed + " seconds.");
}
}

I think that the answer is "probably yes" ... for current Java implementations.
There are two reasons that we can't be sure.
The Java language specification doesn't actually say anything about registers, CPU caches or anything like that. What it actually says is that there is a happens before relationship between one thread writing the volatile and another thread (subsequently) reading it.
While it is reasonable to assume that this will affect caching in the case where there are multiple threads, if the JIT compiler was able to deduce that the volatile variable was thread confined for a given execution of your application, it could reason that it can cache the variable.
That is the theory.
If there was a measurable performance difference, you would be able to measure it in a properly written benchmark. (Though you may get different results depending on the Java version and your hardware.)
However the current version of your benchmark has a number of flaws which would make any results it gives doubtful. If you want to get meaningful results, I strongly recommend that you read the following Q&A.
How do I write a correct micro-benchmark in Java?.
(Unfortunately some of the links in some of the answers seem to be broken ...)

The premise of your benchmark is flawed; the cache is the source of truth. The cache coherence protocol makes sure that CPU caches are coherent; memory is just a spill bucket for whatever doesn't fit into the cache since most caches are write-behind (not write-through). In other words; a write of a volatile doesn't need to be written to main memory; it is sufficient to write it to the cache.
A few examples where writing to the cache isn't desired:
I/O DMA: you want to prevent writing to cache because otherwise, main memory and CPU cache could become incoherent.
Non-temporal data: e.g. you are processing some huge data set and you only access it once, there is no point in caching it.
But this is outside of the reach of a regular Java volatile.
There is a price to pay using volatile:
Atomicity guarantees.
Loads and stores can't be optimized out. This rules out many compiler optimizations. Volatile will not prevent using CPU registers; it will only prevent 'caching' the content of a variable in a register 'indefinitely'.
Ordering guarantees in the form of fences. On the X86 in the above case, the price is at the volatile store. The store needs to wait for the store and all earlier stores to commit to the cache.
Especially the last 2 will impact volatile performance.
Apart from that, your benchmark is flawed. First of all, I would switch to JMH as others already pointed out. It will take care of quite a few typical benchmark errors like warmup and dead code elimination. Also, you should not be using a System.out in the benchmark since this will completely dominate the performance.

Related

Memory barriers on entry and exit of Java synchronized block

I came across answers, here on SO, about Java flushing the work copy of variables within a synchronized block during exit. Similarly it syncs all the variable from main memory once during the entry into the synchronized section.
However, I have some fundamental questions around this:
What if I access mostly non-volatile instance variables inside my synchronized section? Will the JVM automatically cache those variables into the CPU registers at the time of entering into the block and then carry all the necessary computations before finally flushing them back?
I have a synchronized block as below:
The underscored variables _ e.g. _callStartsInLastSecondTracker are all instance variables which I heavily access in this critical section.
public CallCompletion startCall()
{
long currentTime;
Pending pending;
synchronized (_lock)
{
currentTime = _clock.currentTimeMillis();
_tracker.getStatsWithCurrentTime(currentTime);
_callStartCountTotal++;
_tracker._callStartCount++;
if (_callStartsInLastSecondTracker != null)
_callStartsInLastSecondTracker.addCall();
_concurrency++;
if (_concurrency > _tracker._concurrentMax)
{
_tracker._concurrentMax = _concurrency;
}
_lastStartTime = currentTime;
_sumOfOutstandingStartTimes += currentTime;
pending = checkForPending();
}
if (pending != null)
{
pending.deliver();
}
return new CallCompletionImpl(currentTime);
}
Does this mean that all these operations e.g. +=, ++, > etc. requires the JVM to interact with main memory repeatedly? If so, can I use local variables to cache them (preferably stack allocation for primitives) and perform operations and in the end assign them back to the instance variables? Will that help to optimise performance of this block?
I have such blocks in other places as well. On running a JProfiler, it has been observed that most of the time threads are in WAITING state and throughput is also very low. Hence the optimisation necessity.
Appreciate any help here.
(I don't know Java that well, just the underlying locking and memory-ordering concepts that Java is exposing. Some of this is based on assumptions about how Java works, so corrections welcome.)
I'd assume that the JVM can and will optimize them into registers if you access them repeatedly inside the same synchronized block.
i.e. the opening { and closing } are memory barriers (acquiring and releasing the lock), but within that block the normal rules apply.
The normal rules for non-volatile vars are like in C++: the JIT-compiler can keep private copies / temporaries and do full optimization. The closing } makes any assignments visible before marking the lock as released, so any other thread that runs the same synchronized block will see those changes.
But if you read/write those variables outside a synchronized(_lock) block while this synchronized block is executing, there's no ordering guarantee and only whatever atomicity guarantee Java has. Only volatile would force a JVM to re-read a variable on every access.
most of the time threads are in WAITING state and throughput is also very low. Hence the optimisation necessity.
The things you're worried about wouldn't really explain this. Inefficient code-gen inside the critical section would make it take somewhat longer, and that could lead to extra contention.
But there wouldn't be a big enough effect to make most threads be blocked waiting for locks (or I/O?) most of the time, compared to having most threads actively running most of the time.
#Kayaman's comment is most likely correct: this is a design issue, doing too much work inside one big mutex. I don't see loops inside your critical section, but presumably some of those methods you call contain loops or are otherwise expensive, and no other thread can enter this synchronized(_lock) block while one thread is in it.
The theoretical worst case slowdown for store/reload from memory (like compiling C in anti-optimized debug mode) vs. keeping a variable in a register would be for something like while (--shared_var >= 0) {}, giving maybe a 6x slowdown on current x86 hardware. (1 cycle latency for dec eax vs. that plus 5 cycle store-forwarding latency for a memory-destination dec). But that's only if you're looping on a shared var, or otherwise creating a dependency chain through repeated modification of it.
Note that a store buffer with store-forwarding still keeps it local to the CPU core without even having to commit to L1d cache.
In the much more likely case of code that just reads a var multiple times, anti-optimized code that really loads every time can have all those loads hit in L1d cache very efficiently. On x86 you'd probably barely notice the difference, with modern CPUs having 2/clock load throughput, and efficient handling of ALU instructions with memory source operands, like cmp eax, [rdi] being basically as efficient as cmp eax, edx.
(CPUs have coherent caches so there's no need for flushing or going all the way to DRAM to ensure you "see" data from other cores; a JVM or C compiler only has to make sure the load or store actually happens in asm, not optimized into a register. Registers are thread-private.)
But as I said, there's no reason to expect that your JVM is doing this anti-optimization inside synchronized blocks. But even if it were, it might make a 25% slowdown.
You are accessing members on a single object. So when the CPU reads the _lock member, it needs to load the cache line containing _lock member first. So probably quite a few of the member variables will be on the same cache line which is already in your cache.
I would be more worried about the synchronized block itself IF you have determined it is actually a problem; it might not be a problem at all. For example Java uses quite a few lock optimization techniques like biased locking, adaptive spin lock to reduce the costs of locks.
But if it is a contended lock, you might want to make the duration of the lock shorter by moving as much out of the lock as possible and perhaps even get rid of the whole lock and switch to a lock free approach.
I would not trust JPofiler for a second.
http://psy-lob-saw.blogspot.com/2016/02/why-most-sampling-java-profilers-are.html
So it might be that JProfiler is putting you in the wrong direction.

Usage of lazySet on AtomicXXX in Java

From this question : AtomicInteger lazySet vs. set and form this link : https://docs.oracle.com/javase/7/docs/api/java/util/concurrent/atomic/package-summary.html
I could gather following points
lazySet could be faster than set
lazySet uses store-store barrier (writes before are honored but not the contended writes, which were yet to happen)
I could find one use-case where it could be applied, from the documentation :
Use lazySet when you want to null out a pointer to aid GC.
Are there any other practical use-cases for lazySet ?
Caffeine uses lazy or relaxed writes in many of its data structures.
When nulling out a field (e.g. ConcurrentLinkedStack)
When writing to volatile fields before publishing (e.g. SingleConsumerQueue)
When publish is safely delayable (e.g. BoundedBuffer)
When races are benign (e.g. cache expiration timestamps)
When inside a lock (e.g. BoundedLocalCache)
ConcurrentLinkedQueue uses relaxed writes prior to publishing a node and may lazily sets a node's next field (prior to publishing or to indicate a stale traversal).
You may also enjoy reading the Linux Kernel Memory Barriers paper.
TL;DR How to use .lazySet()? With care, if at all.
The main problem here is that AtomicXXX.lazySet() is low-level performance optimization and it is out of current JLS. You can't prove correctness if your concurrent code with JMM tools if you are using lazySet().
Why it is much faster than volatile write?
Main difference between set and lazySet is absence of StoreLoad barrier.
JSR-133 Cookbook for Compiler Writers:
StoreLoad barriers are needed on nearly all recent multiprocessors, and are usually the most expensive kind.
Moreover, on most popular x86-based hardware StoreLoad is the only explicit barrier (others are just no-op's and cost nothing), so with lazySet you eliminate all (explicit) memory barriers.
Guarantees of lazySet
From the point of JLS there isn't any.
But actually you can reason about lazySet as delayed write which cannot be reordered with any previous write and will happen eventually. Eventually is finite time, if your process makes any progress (e.g. any synchronization action occurs; in addition, size of processor's store buffer is finite). If written value became visible for other thread, you can be sure that all previous writes are visible either (although you cannot formally prove it). So you can treat it as delayed happens-before relationship (but, of course, it's not even close to it's strict and formal definition).
Usage
Most practical usage (except nulling-out references) is making writes far cheaper in context of progress. Simplest example is using lazySet() instead of set() within synchronized block (but in this case there is no great performance impact). Or you can use it instead of writes in single producer cases, when no compete on write occurs.
Disruptor developers are using lazySet exactly for this purpose in their lock-free implementation. Again, it's very hard to argue about correctness of such code, but it's good trick to be aware of.
I would think many uses of AtomicBoolean would benefit from the usage of lazySet() because they are often used as flags to indicate whether something is complete or not, or an outer loop should finish.
This is because in this case the value is initially one value and it eventually becomes another value and then stays there. Obviously this argument applied to almost any atomic that is used in that way.
public void test() {
final AtomicBoolean finished = new AtomicBoolean(false);
new Thread(new Runnable() {
#Override
public void run() {
while (!finished.get()) {
// A long process.
if (wereAllDone()) {
finished.lazySet(true);
}
}
}
}).start();
}

Increased cost of a volatile write over a nonvolatile write

I've been reading about volatile (https://www.ibm.com/developerworks/java/library/j-jtp06197/) and came across a bit that says that a volatile write is so much more expensive than a nonvolatile write.
I can understand that there would be an increased cost associated with a volatile write given that volatile is a way of synchronization but want to know how exactly how a volatile write is so much more expensive than a nonvolatile write; does it perhaps have to do with visibility across different thread stacks at the time at which the volatile write is made?
Here's why, according to the article you have indicated:
Volatile writes are considerably more expensive than nonvolatile writes because of the memory fencing required to guarantee visibility but still generally cheaper than lock acquisition.
[...] volatile reads are cheap -- nearly as cheap as nonvolatile reads
And that is, of course, true: memory fence operations are always bound to writing and reads execute the same way regardless of whether the underlying variable is volatile or not.
However, volatile in Java is about much more than just volatile vs. nonvolatile memory read. In fact, in its essence it has nothing to do with that distinction: the difference is in the concurrent semantics.
Consider this notorious example:
volatile boolean runningFlag = true;
void run() {
while (runningFlag) { do work; }
}
If runningFlag wasn't volatile, the JIT compiler could essentially rewrite that code to
void run() {
if (runningFlag) while (true) { do work; }
}
The ratio of overhead introduced by reading the runningFlag on each iteration against not reading it at all is, needless to say, enormous.
It is about caching. Since new processors use caches, if you don't specify volatile data stays in cache and operation of writing is fast. (Since cache is near processor) If variable is marked as volatile, system needs to write it fully into memory nad that is a bit slower operation.
And yes you are thinking right it has to do something with different thread stacks, since each is separate and reads from SAME memory, but not necessarily from same cache. Today processors use many levels of caching so this can be a big problem if multiple threads/processes are using same data.
EDIT: If data stays in local cache other threads/processes won't see change until data is written back in memory.
Most likely it has to do with the fact that a volatile write has to stall the pipeline.
All writes are queued to be written to the caches. You don't see this with non-volatile writes/reads as the code can just get the value you just wrote without involving the cache.
When you use a volatile read, it has to go back to the cache, and this means the write (as implemented) cannot continue under the write has been written to the case (in case you do a write followed by a read)
One way around this is to use a lazy write e.g. AtomicInteger.lazySet() which can be 10x faster than a volatile write as it doesn't wait.

Are volatile variable 'reads' as fast as normal reads?

I know that writing to a volatile variable flushes it from the memory of all the cpus, however I want to know if reads to a volatile variable are as fast as normal reads?
Can volatile variables ever be placed in the cpu cache or is it always fetched from the main memory?
You should really check out this article: http://brooker.co.za/blog/2012/09/10/volatile.html. The blog article argues volatile reads can be a lot slower (also for x86) than non-volatile reads on x86.
Test 1 is a parallel read and write to a non-volatile variable. There
is no visibility mechanism and the results of the reads are
potentially stale.
Test 2 is a parallel read and write to a volatile variable. This does not address the OP's question specifically. However worth noting that a contended volatile can be very slow.
Test 3 is a read to a volatile in a tight loop. Demonstrated is that the semantics of what it means to be volatile indicate that the value can change with each loop iteration. Thus the JVM can not optimize the read and hoist it out of the loop. In Test 1, it is likely the value was read and stored once, thus there is no actual "read" occurring.
Credit to Marc Booker for running these tests.
The answer is somewhat architecture dependent. On an x86, there is no additional overhead associated with volatile reads specifically, though there are implications for other optimizations.
JMM cookbook from Doug Lea, see architecture table near the bottom.
To clarify: There is not any additional overhead associated with the read itself. Memory barriers are used to ensure proper ordering. JSR-133 classifies four barriers "LoadLoad, LoadStore, StoreLoad, and StoreStore". Depending on the architecture, some of these barriers correspond to a "no-op", meaning no action is taken, others require a fence. There is no implicit cost associated with the Load itself, though one may be incurred if a fence is in place. In the case of the x86, only a StoreLoad barrier results in a fence.
As pointed out in a blog post, the fact that the variable is volatile means there are assumptions about the nature of the variable that can no longer be made and some compiler optimizations would not be applied to a volatile.
Volatile is not something that should be used glibly, but it should also not be feared. There are plenty of cases where a volatile will suffice in place of more heavy handed locking.
It is architecture dependent. What volatile does is tell the compiler not to optimise that variable away. It forces most operations to treat the variable's state as an unknown. Because it is volatile, it could be changed by another thread or some other hardware operation. So, reads will need to re-read the variable and operations will be of the read-modify-write kind.
This kind of variable is used for device drivers and also for synchronisation with in-memory mutexes/semaphores.
Volatile reads cannot be as quick, especially on multi-core CPUs (but also only single-core).
The executing core has to fetch from the actual memory address to make sure it gets the current value - the variable indeed cannot be cached.
As opposed to one other answer here, volatile variables are not used just for device drivers! They are sometimes essential for writing high performance multi-threaded code!
volatile implies that the compiler cannot optimize the variable by placing its value in a CPU register. It must be accessed from main memory. It may, however, be placed in a CPU cache. The cache will guaranty consistency between any other CPUs/cores in the system. If the memory is mapped to IO, then things are a little more complicated. If it was designed as such, the hardware will prevent that address space from being cached and all accesses to that memory will go to the hardware. If there isn't such a design, the hardware designers may require extra CPU instructions to insure that the read/write goes through the caches, etc.
Typically, the 'volatile' keyword is only used for device drivers in operating systems.

Java synchronization and performance in an aspect

I just realized that I need to synchronize a significant amount of data collection code in an aspect but performance is a real concern. If performance degrades too much my tool will be thrown out. I will be writing ints and longs individually and to various arrays, ArrayLists and Maps. There will be multiple threads of an application that will make function calls that will be picked up by my aspect. What kind of things should I look out for that will negatively affect performance? What code patterns are more efficient?
In particular I have a method that calls many other data recording methods:
void foo() {
bar();
woz();
...
}
The methods mostly do adding an incrementing of aspect fields
void bar() {
f++; // f is a field of the aspect
for (int i = 0; i < ary.length; i++) {
// get some values from aspect point cut
if (some condiction) {
ary[i] += someValue; // ary a field of the aspect
}
}
}
Should I synchronize foo, or bar, woz and others individually, or should I move all the code in bar, woz, etc into foo and just synchronize it? Should I synchronize on this, on a specifically created synchronization object:
private final Object syncObject = new Object();
(see this post), or on individual data elements within the methods:
ArrayList<Integer> a = new ArrayList<Integer>();
void bar() {
synchronize(a) {
// synchronized code
}
}
Concurrency is extremely tricky. It's very easy to get it wrong, and very hard to get right. I wouldn't be too terribly worried about performance at this point. My first and foremost concern would be to get the concurrent code to work safely (no deadlocks or race conditions).
But on the issue of performance: when in doubt, profile. It's hard to say just how different synchronization schemes will affect performance. It's even harder for us to give you suggestions. We'd need to see a lot more of your code and gain a much deeper understanding of what the application does to give you a truly useful answer. In contrast, profiling gives you hard evidence as to if one approach is slower than another. It can even help you identify where the slowdown is.
There are a lot of great profiling tools for Java these days. The Netbeans and Eclipse profilers are good.
Also, I'd recommend staying away from raw synchronization altogether. Try using some of the classes in the java.util.concurrency package. They make writing concurrent code much easier, and much less error prone.
Also, I recommend you read Java Concurrency in Practice by Brian Goetz, et al. It's very well written and covers a lot of ground.
Rule of thumb is not to synchronize on this - most of the times it is a performance hit - all methods are synchronized on one object.
Consider using locks - they'a very nice abstraction and many fine features like, trying to lock for a time period, and then giving up:
if(commandsLock.tryLock(100, TimeUnit.MILLISECONDS)){
try {
//Do something
}finally{
commandsLock.unlock();
}
}else{
//couldnt acquire lock for 100 ms
}
I second opinion on using java.util.concurrent. I'd make two levls of synchronization
synchronize collection access (if it is needed)
synchronize field access
Collection access
If your collection are read-only ie no elements get removed-inserted (but elements may change) i would say that you should use synchronized collections (but this may be not needed...) and dont synchronize iterations:
Read only:
for (int i = 0; i < ary.length; i++) {
// get some values from aspect point cut
if (some condiction) {
ary += someValue; // ary a field of the aspect
}
}
and ary is instance obtained by Collections.synchronizedList.
Read-write
synchronized(ary){
for (int i = 0; i < ary.length; i++) {
// get some values from aspect point cut
if (some condiction) {
ary += someValue; // ary a field of the aspect
}
}
}
Or use some concurrent collections (like CopyOnWriteArrayList) which is inherentently therad safe.
Main difference is that - in first read-only wersion any number of threads may iterate over this collections, and in second only one at a time may iterate. In both cases only one therad at a time should increment any given field.
Field access
Synchronize incrementations on fields separately from synchronizing iterations.
like:
Integer foo = ary.get(ii);
synchronized(foo){
foo++;
}
Get rid of synchronization
Use concurrent collections (from java.util.concurrent - not from `Collections.synchronizedXXX', latter still need synchronizing on traversal).
Use java.util.atomic that enable you to atomically incrememt fields.
Something you should watch:
Java memory model - its a talk that gives very nice understanding on how synchronizations and data aligment in JAVA works.
Upadte: since writing the below, I see you've updated the question slightly. Forgive my ignorance-- I have no idea what an "aspect" is-- but from the sample code you posted, you could also consider using atomics/concurrent collections (e.g. AtomicInteger, AtomicIntegerArray) or atomic field updaters. This could mean quite a re-factoring of your code, though. (In Java 5 on a dual-proc hyperthreading Xeon, the throughput of AtomicIntegerArray is significantly better than a synchronized array; sorry, I haven't got round to repeating the test on more procs/later JVM version yet-- note that performance of 'synchronized' has improved since then.)
Without more specific information or metrics about your particular program, the best you can do is just follow good program design. It's worth noting that the performance and optimisation of synchronization locks in the JVM has beed one of the areas (if not, the area) that has received most research and attention over the last few years. And so in the latest versions of JVM's, it ain't all that bad.
So in general, I'd say synchronize minimally without "going mad". By 'minimally', I mean so that you hold on to the lock for as less time as possible, and so that only the parts that need to use that specific lock use that specific lock. But only if the change is easy to do and it's easy to prove that your program is still correct. For example, instead of doing this:
synchronized (a) {
doSomethingWith(a);
longMethodNothingToDoWithA();
doSomethingWith(a);
}
consider doing this if and only if your program will still be correct:
synchronized (a) {
doSomethingWith(a);
}
longMethodNothingToDoWithA();
synchronized (a) {
doSomethingWith(a);
}
But remember, the odd simple field update with a lock held unnecessarily probably won't make much tangible difference, and could actually improve performance. Sometimes, holding a lock for a bit longer and doing less lock "housekeeping" can be beneficial. But the JVM can make some of those decisions, so you don't need to be tooo paranoid-- just do generally sensible things and you should be fine.
In general, try and have a separate lock for each set of methods/accesses that together form an "independent process". Other than that, having a separate lock object can be a good way of encapsulating the lock within the class it's used by (i.e. preventing it from being used by outside callers in a way you didn't predict), but there's probably no performance difference per se from using one object to another as the lock (e.g. using the instance itself vs a private Object declared just to be a lock within that class as you suggest), provided the two objects would otherwise be used in exactly the same way.
There should be a performance difference between a built-in language construct and a library, but experience has taught me not to guess when it comes to performance.
If you compile the aspect into the application then you will have basically no performance hit, if you do it at runtime (load-type weaving) then you will see a performance hit.
If you have each aspect be perinstance then it may reduce the need for synchronization.
You should have as little synchronization as possible, for as short a time as possible, to reduce any problems.
If possible you may want to share as little state as possible between threads, keeping as much local as possible, to reduce any deadlock problems.
More information would lead to a better answer btw. :)

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