Speed issues with ReaderWriterLockSlim and Garbage Collection - garbage-collection

I have an example piece of code the illustrates issues in my code when GC.Collect is carried out on a class having a ReaderWriterLockSlim member variable. The GC.Collect takes between 2 and 3 seconds to run. I need to carry out GC at regular intervals because my applicaton is extremely memory intensive.
namespace WpfApplication12
{
public class DataItem
{
private readonly ReaderWriterLockSlim m_propertyLock = new ReaderWriterLockSlim();
public DataItem()
{
}
}
/// <summary>
/// Interaction logic for Window1.xaml
/// </summary>
public partial class Window1 : Window
{
public Window1()
{
InitializeComponent();
List<DataItem> dataItemList = new List<DataItem>();
for (int i = 0; i < 100000; i++)
{
dataItemList.Add(new DataItem());
}
Debug.WriteLine(DateTime.Now.ToString());
GC.Collect();
Debug.WriteLine(DateTime.Now.ToString());
}
}
}
Has anyone had similar problems?
Thanks
Ian

I'd ask if you really need a ReaderWriterLockSlim for each of your DataItem classes?
Seems like bad design to me to have that many handles floating about. After-all, that's what will be causing the delay...

The memory issue may be caused if the readerwriterlockslim is called from multiple threads. I believe it will store some information of the threads which can cause the memory consumption to bloat. I would recommend trying to figure out a solution where you can bring down the number of thread that are calling the readerwriterlockslim.

Related

does dispose method disposes the calling object also?

I found the following code on MSDN:
public class DisposeExample
{
public class MyResource: IDisposable
{
private IntPtr handle;
private Component component = new Component();
private bool disposed = false;
public MyResource(IntPtr handle)
{
this.handle = handle;
}
public void Dispose()
{
Dispose(true);
GC.SuppressFinalize(this);
}
protected virtual void Dispose(bool disposing)
{
if(!this.disposed)
{
if(disposing)
{
// Dispose managed resources.
component.Dispose();
}
CloseHandle(handle);
handle = IntPtr.Zero;
disposed = true;
}
}
~MyResource()
{
Dispose(false);
}
}
public static void Main()
{
MyResource obj = new MyResource()
//obj.dispose()
}
}
Now the confusion I have here is that, if I call obj.dispose, it disposes the objects created in the class MyResources i.e. handle, component etc. But does the obj also gets removed off the heap?? Same applies with the destructor. If I don't call dispose, the destructor will be called sometime. The code inside destructor removes the contained objects. But what about the obj?
Secondly, if I don't have a destructor defined inside the class and I dont even call dispose, does the GC never come into picture here?
IDisposable exists to remove unmanaged items from your managed objects. The runtime automatically provides a destructor, this destructor here has the sole purpose of releasing unmanaged items. As soon as your object goes out of scope or is set to null and has no more references to it will eventually be cleared by the GC.
The fundamental rule I'd recommend with with IDisposable is that at any given moment in time, for every object that implements IDisposable, there should be exactly one entity which has the clearly-defined responsibility of ensuring that it will get cleaned up (almost always by calling Dispose) before it is abandoned. Such responsibility will initially belong to whatever entity calls the IDidposable object's constructor, but that entity may hand the responsibility off to some other entity (which may hand it off again, etc.).
In general, I'd guess that most programmers would be best served if they pretended finalizers and destructors did not exist. They are generally only needed as a consequence of poorly-written code, and in most cases the effort one would have to spend writing a 100%-correct finalizer/destructor and working through all the tricky issues related to threading context, accidental resurrection, etc. could be better spent ensuring that the primary rule given above is always followed. Because of some unfortunate design decisions in the Framework and its languages, there are a few situations which can't very well be handled without finalizers/destructors, but for the most part they serve to turn code which would fail relatively quickly into code which will mostly work but may sometimes fail in ways that are almost impossible to debug.

Thread-Safe Properties in C#

I know that this subject is slightly "Played Out", but I am still terribly confused. I have a class with properties that will be updates by multiple threads and I am trying to allow the properties to be updated in a Threadsafe manner.
Below, I have included a few examples of what I have tried thus far (the class is contained within a BindingList so its properties call a PropertyChangingEventHandler event).
Method 1 - Doubles
private double _Beta;
public double Beta
{
get
{
return _Beta;
}
}
private readonly BetaLocker = new object();
public void UpdateBeta(double Value)
{
lock (BetaLocker)
{
_Beta = Value;
NotifyPropertyChanged("Beta");
}
}
Method 2 - Ints
private int _CurrentPosition;
public int CurrentPosition
{
get
{
return _CurrentPosition;
}
}
public void UpdatePosition(int UpdateQuantity)
{
Interlocked.Add(ref _CurrentPosition, UpdateQuantity);
NotifyPropertyChanged("CurrentPosition");
}
Basically - is the current way that I am creating properties completely threadsafe for both ints and doubles?
You have to ask yourself what it means to be Thread Safe (yes, it's a link to wikipedia and it's blacked out ^_^):
A piece of code is thread-safe if it only manipulates shared data structures in a manner that guarantees safe execution by multiple threads at the same time. There are various strategies for making thread-safe data structure
So now you have to determine if your code guarantees safe execution if executed by multiple threads: the quick answer is that both of your code samples are thread safe! However (and this is a big one), you also have to consider the usage of the object and determine if it is Thread Safe also... here is an example:
if(instance.Beta==10.0)
{
instance.UpdateBeta(instance.Beta*10.0);
}
// what's instance.Beta now?
In this case you have absolutely no guarantee that Beta will be 100.0 because beta could have changed after you checked it. Imagine this situation:
Thread 2: UpdateBeta(10.0)
Thread 1: if(Beta == 10.00)
Thread 2: UpdateBeta(20.0)
Thread 1: UpdateBeta(Beta*10.0)
// Beta is now 200.0!!!
The quick and dirty way to fix this is to use a double-checked lock:
if(instance.Beta==10.0)
{
lock(instance)
{
if(instance.Beta==10.0)
{
instance.UpdateBeta(instance.Beta*10.0);
}
}
}
The same is true for CurrentPosition.

IDisposable + finalizer pattern

Looking at the IDisposable pattern + Finalizer pattern, there is something I don't understand:
public class ComplexCleanupBase : IDisposable
{
private bool disposed = false; // to detect redundant calls
public ComplexCleanupBase()
{
// allocate resources
}
protected virtual void Dispose(bool disposing)
{
if (!disposed)
{
if (disposing)
{
// dispose-only, i.e. non-finalizable logic
}
// shared cleanup logic
disposed = true;
}
}
~ComplexCleanupBase()
{
Dispose(false);
}
public void Dispose()
{
Dispose(true);
GC.SuppressFinalize(this);
}
}
From my understanding the pattern should be implemented like above.
1) Calling Dispose() fires GC.SuppressFinalize(this), which means that the object should not be put on the finalizer queue as its already properly disposed? That helps to free up the object faster?
2) But what if I am not calling Dispose() on this object at all? In that case the finalizer should kick in, correct? But Dispose(false); does absolutely nothing (only setting disposed = true). Is this intended? It feels like as if something is missing...
Question 1: Yes, if GC.SuppressFinalize is not called the object will be placed on the finalizer queue AND will move up a generation (if not already gen 2), which means that the memory for that object will not be reclaimed until the next pass of GC for the new generation.
Question 2: Your comment //shared cleanup logic is where the shared cleanup logic will go, this is where things other than setting disposed = true would happen.
Also, an aside: if your dispose logic should only be called once, consider acquiring a lock, an uncontested lock is very fast in .Net:
public class ComplexCleanupBase : IDisposable
{
private volatile bool disposed = false; // to detect redundant calls
private readonly object _mutex = new object();
protected virtual void Dispose(bool disposing)
{
if (!Monitor.TryEnter(_mutex))
{
// We're already being disposed.
return;
}
try
{
if (!disposed)
{
if (disposing)
{
// dispose-only, i.e. non-finalizable logic
}
// shared cleanup logic
disposed = true;
}
}
finally
{
Monitor.Exit(_mutex);
}
}
... other methods unchanged
}
If Dispose(false) isn't going to do anything, that's a very good indication that neither your class nor any class derived from it should include a C#-style "destructor" nor override Finalize, and the "disposing" argument should be regarded as a dummy whose purpose is to give the protected Dispose method a different signature from the public one.
Note that implementing a destructor or overriding Finalize in a derived class, when the parent class is not expecting such behavior, can produce Heisenbugs. Among other things, the GC can sometimes decide that a class object has been abandoned, triggering its finalizer/destructor, even while an entity referred to by a field of the class is being used. For example, suppose a static class usbThingie manipulates USB controllers using integer handles, and a wrapper class usbWrapper does something like:
UInt32 myHandle;
void sendData(Byte data[])
{
UsbThingie.send(myHandle, data[0], data.Length);
}
If a sendData() call is the last thing done to an instance of usbWrapper before it is abandoned, it would be possible for the garbage-collector to observe that once UsbThingie.send() is called--even before it returns--no further references will exist to the usbWrapper, and thus it can safely trigger the finalizer. If the finalizer tries to close the channel referred to by myHandle, that might disrupt the transmission that was taking place; if usbThingie isn't thread-safe, there's no telling what might happen.

java:singleton, static variable and thread safety

class MyClass
{
private static MyClass obj;
public static MyClass getInstance()
{
if(obj==null)
{
obj = new MyClass();
}
return obj;
}
In the above java code sample, because obj is a static variable inside the class,
will getInstance still be non-thread safe? Because static variables are shared by all threads, 2 simultaneous threads shall be using the same object. Isnt it?
Vipul Shah
Because static variables are so widely shared they are extremely un-thread safe.
Consider what happens if two threads call your getInstance at the same time. Both threads will be looking at the shared static obj and both threads will see that obj is null in the if check. Both threads will then create a new obj.
You may think: "hey, it is thread safe since obj will only ever have one value, even if it is initialized multiple times." There are several problems with that statement. In our previous example, the callers of getInstance will both get their own obj back. If both callers keep their references to obj then you will have multiple instances of your singleton being used.
Even if the callers in our previous example just did: MyClass.getInstance(); and didn't save a reference to what MyClass.getInstance(); returned, you can still end up getting different instances back from getInstance on those threads. You can even get into the condition where new instances of obj are created even when the calls to getInstance do not happen concurrently!
I know my last claim seems counter-intuitive since the last assignment to obj would seem to be the only value that could be returned from future calls to MyClass.getInstance(). You need to remember, however, that each thread in the JVM has its own local cache of main memory. If two threads call getInstance, their local caches could have different values assigned to obj and future calls to getInstance from those threads will return what is in their caches.
The simplest way to make sure that getInstance thread safe would be to make the method synchronized. This will ensure that
Two threads can not enter getInstance at the same time
Threads trying to use obj will never get a stale value of obj from their cache
Don't try to get clever and use double checked locking:
http://www.cs.umd.edu/~pugh/java/memoryModel/DoubleCheckedLocking.html
Good explanation can be found here:
http://en.wikipedia.org/wiki/Singleton_pattern
The wiki article highlights various thread-safe approaches along with some of their pros and cons.
in this case getInstance() is not thread-safe, even if you use static variable. only synchronization makes this thread-safe.
The following example shows a weird thread save modified single ton pattern which supports generics as well.
To have it just thread save and synchronization save just take the synchronized block and the transient and volatile keywords.
Notice, that there is a double check, the synchronized block is inside an if. This brings more performance, because synchronized is expensive.
Of course for a real singleton do not use maps, I said it is a modified one.
public class Edge<T> {
#SuppressWarnings({"unchecked"})
private static transient volatile HashMap<Object,HashMap<Object, Edge>> instances = new HashMap<Object, HashMap<Object,Edge>>();
/**
* This function is used to get an Edge instance
* #param <T> Datatype of the nodes.
* #param node1, the source node
* #param node2, the destination node
* #return the edge of the two nodes.
*/
#SuppressWarnings({"unchecked"})
public static <T> Edge<T> getInstance(T node1, T node2){
if(!(instances.containsKey(node1) && instances.get(node1).containsKey(node2))){
synchronized (Edge.class) {
if(!(instances.containsKey(node1) && instances.get(node1).containsKey(node2))){
Edge<T> edge = new Edge<T>(node1, node2);
if(!instances.containsKey(node1)){
instances.put(node1, new HashMap<Object, Edge>());
}
instances.get(node1).put(node2, edge);
}
}
}
return (Edge<T>)instances.get(node1).get(node2);
}
public class Singleton{
private static transient volatile Singleton instance;
public static Singleton getInstance(){
if(instance==null)synchronized(Singleton.class){
if(instance==null){
instance = new Singleton();
}
}
return instance;
}
private Singleton(){
/*....*/
}
}
Page 182:
http://books.google.com/books?id=GGpXN9SMELMC&printsec=frontcover&dq=design+patterns&hl=de&ei=EFGCTbyaIozKswbHyaiCAw&sa=X&oi=book_result&ct=result&resnum=2&ved=0CDMQ6AEwAQ#v=onepage&q&f=false
Think this can be tagged as answered now.
class MyClass
{
private static MyClass obj;
private MyClass(){
// your initialization code
}
public static synchronized MyClass getInstance()
{
if(obj==null)
{
obj = new MyClass();
}
return obj;
}
I'll agree with #Manoj.
I believe the above will be one of the best methods to achieve singleton object.
And synchronization makes the object thread safe.
Even, it's static :)

Threading from within a class with static and non-static methods

Let's say I have
class classA {
void someMethod()
{
Thread a = new Thread(threadMethod);
Thread b = new Thread(threadMethod);
a.Start();
b.Start();
a.Join();
b.Join();
}
void threadMethod()
{
int a = 0;
a++;
Console.Writeline(a);
}
}
class classB {
void someMethod()
{
Thread a = new Thread(threadMethod);
Thread b = new Thread(threadMethod);
a.Start();
b.Start();
a.Join();
b.Join();
}
static void threadMethod()
{
int a = 0;
a++;
Console.Writeline(a);
}
}
Assuming that in classA and classB, the contents of threadMethod have no effect to anything outside of its inner scope, does making threadMethod in classB static have any functional difference?
Also, I start two threads that use the same method in the same class. Does each method get its own stack and they are isolated from one another in both classA and classB?
Does again the static really change nothing in this case?
Methods don't have stacks, threads do. In your example threadMethod only uses local variables which are always private to the thread executing the method. It doesn't make any difference if the method is static or not as the method isn't sharing any data.
In this case there is no functional difference. Each thread gets it's own stack
Maybe you can be a little more clear. It doesn't matter if the function is declared static or not in most languages. Each thread has its own private statck.
Each thread would get it's own stack. There is no functional difference that I can tell between the two.
The only difference (obviously) is that the static version would be unable to access member functions/variables.

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