I known a Hashtable is synchronized, but why its get() method is synchronized?
Is it only a read method?
If the read was not synchronized, then the Hashtable could be modified during the execution of read. New elements could be added, the underlying array could become too small and could be replaced by a bigger one, etc. Without sequential execution, it is difficult to deal with these situations.
However, even if get would not crash when the Hashtable is modified by another thread, there is another important aspect of the synchronized keyword, namely cache synchronization. Let's use a simplified example:
class Flag {
bool value;
bool get() { return value; } // WARNING: not synchronized
synchronized void set(bool value) { this->value = value; }
}
set is synchronized, but get isn't. What happens if two threads A and B simultaneously read and write to this class?
1. A calls read
2. B calls set
3. A calls read
Is it guaranteed at step 3 that A sees the modification of thread B?
No, it isn't, as A could be running on a different core, which uses a separate cache where the old value is still present. Thus, we have to force B to communicate the memory to other core, and force A to fetch the new data.
How can we enforce it? Everytime, a thread enters and leaves a synchronized block, an implicit memory barrier is executed. A memory barrier forces the cache to be updated. However, it is required that both the writer and the reader have to execute the memory barrier. Otherwise, the information is not properly communicated.
In our example, thread B already uses the synchronized method set, so its data modification is communicated at the end of the method. However, A does not see the modified data. The solution is to make get synchronized, so it is forced to get the updated data.
Have a look in Hashtable source code and you can think of lots of race conditions that can cause problem in a unsynchronized get() .
(I am reading JDK6 source code)
For example, a rehash() will create a empty array, and assign it to the instance var table, and put the entries from old table to the new one. Therefore if your get occurs after the empty array assignment, but before actually putting entries in it, you cannot find your key even it is in the table.
Another example is, there is a loop iterate thru the linked list at the table index, if in middle in your iteration, rehash happens. You may also failed to find the entry even it exists in the hashtable.
Hashtable is synchronized meaning the whole class is thread-safe
Inside the Hashtable, not only get() method is synchronized but also many other methods are. And particularly put() method is synchronized like Tom said.
A read method must be synchronized as a write method because it will make sure the visibility and the consistency of the variable.
Related
A bit of (simplified) context.
Let's say I have an ArrayList<ContentStub> where ContentStub is:
public class ContentStub {
ContentType contentType;
Object content;
}
And I have multiple implementations of classes that "inflate" stubs for each ContentType, e.g.
public class TypeAStubInflater {
public void inflate(List<ContentStub> contentStubs) {
contentStubs.forEach(stub ->
{
if(stub.contentType == ContentType.TYPE_A) {
stub.content = someService.getContent();
}
});
}
}
The idea being, there is TypeAStubInflater which only modifies items ContentType.TYPE_A running in one thread, and TypeBStubInflater which only modifies items ContentType.TYPE_B, etc. - but each instance's inflate() method is modifying items in the same contentStubs List, in parallel.
However:
No thread ever changes the size of the ArrayList
No thread ever attempts to modify a value that's being modified by another thread
No thread ever attempts to read a value written by another thread
Given all this, it seems that no additional measures to ensure thread-safety are necessary. From a (very) quick look at the ArrayList implementation, it seems that there is no risk of a ConcurrentModificationException - however, that doesn't mean that something else can't go wrong. Am I missing something, or this safe to do?
In general, that will work, because you are not modifying the state of the List itself, which would throw a ConcurrentModificationException if any iterator is active at the time of looping, but rather are modifying just an object inside the list, which is fine from the list's POV.
I would recommend splitting up your into a Map<ContentType, List<ContentStub>> and then start Threads with those specific lists.
You could convert your list to a map with this:
Map<ContentType, ContentStub> typeToStubMap = stubs.stream().collect(Collectors.toMap(stub -> stub.contentType, Function.identity()));
If your List is not that big (<1000 entries) I would even recommend not using any threading, but just use a plain for-i loop to iterate, even .foreach if that 2 extra integers are no concern.
Let's assume the thread A writes TYPE_A content and thread B writes TYPE_B content. The List contentStubs is only used to obtain instances of ContentStub: read-access only. So from the perspective of A, B and contentStubs, there is no problem. However, the updates done by threads A and B will likely never be seen by another thread, e.g. another thread C will likely conclude that stub.content == null for all elements in the list.
The reason for this is the Java Memory Model. If you don't use constructs like locks, synchronization, volatile and atomic variables, the memory model gives no guarantee if and when modifications of an object by one thread are visible for another thread. To make this a little more practical, let's have an example.
Imagine that a thread A executes the following code:
stub.content = someService.getContent(); // happens to be element[17]
List element 17 is a reference to a ContentStub object on the global heap. The VM is allowed to make a private thread copy of that object. All subsequent access to reference in thread A, uses the copy. The VM is free to decide when and if to update the original object on the global heap.
Now imagine a thread C that executes the following code:
ContentStub stub = contentStubs.get(17);
The VM will likely do the same trick with a private copy in thread C.
If thread C already accessed the object before thread A updated it, thread C will likely use the – not updated – copy and ignore the global original for a long time. But even if thread C accesses the object for the first time after thread A updated it, there is no guarantee that the changes in the private copy of thread A already ended up in the global heap.
In short: without a lock or synchronization, thread C will almost certainly only read null values in each stub.content.
The reason for this memory model is performance. On modern hardware, there is a trade-off between performance and consistency across all CPUs/cores. If the memory model of a modern language requires consistency, that is very hard to guarantee on all hardware and it will likely impact performance too much. Modern languages therefore embrace low consistency and offer the developer explicit constructs to enforce it when needed. In combination with instruction reordering by both compilers and processors, that makes old-fashioned linear reasoning about your program code … interesting.
I have been trying to figure out that how immutable objects which are safely published could be observed with stale reference.
public final class Helper {
private final int n;
public Helper(int n) {
this.n = n;
}
}
class Foo {
private Helper helper;
public Helper getHelper() {
return helper;
}
public void setHelper(int num) {
helper = new Helper(num);
}
}
So far I could understand that Helper is immutable and can be safely published. A reading thread either reads null or fully initialized Helper object as it won't be available until fully constructed. The solution is to put volatile in Foo class which I don't understand.
The fact that you are publishing a reference to an immutable object is irrelevant here.
If you are reading the value of a reference from multiple threads, you need to ensure that the write happens before a read if you care about all threads using the most up-to-date value.
Happens before is a precisely-defined term in the language spec, specifically the part about the Java Memory Model, which allows threads to make optimisations for example by not always updating things in main memory (which is slow), instead holding them in their local cache (which is much faster, but can lead to threads holding different values for the "same" variable). Happens-before is a relation that helps you to reason about how multiple threads interact when using these optimisations.
Unless you actually create a happens-before relationship, there is no guarantee that you will see the most recent value. In the code you have shown, there is no such relationship between writes and reads of helper, so your threads are not guaranteed to see "new" values of helper. They might, but they likely won't.
The easiest way to make sure that the write happens before the read would be to make the helper member variable final: the writes to values of final fields are guaranteed to happen before the end of the constructor, so all threads always see the correct value of the field (provided this wasn't leaked in the constructor).
Making it final isn't an option here, apparently, because you have a setter. So you have to employ some other mechanism.
Taking the code at face value, the simplest option would be to use a (final) AtomicInteger instead of the Helper class: writes to AtomicInteger are guaranteed to happen before subsequent reads. But I guess your actual helper class is probably more complicated.
So, you have to create that happens-before relationship yourself. Three mechanisms for this are:
Using AtomicReference<Helper>: this has similar semantics to AtomicInteger, but allows you to store a reference-typed value. (Thanks for pointing this out, #Thilo).
Making the field volatile: this guarantees visibility of the most recently-written value, because it causes writes to flush to main memory (as opposed to reading from a thread's cache), and reads to read from main memory. It effectively stops the JVM making this particular optimization.
Accessing the field in a synchronized block. The easiest thing to do would be to make the getter and setter methods synchronized. Significantly, you should not synchronize on helper, since this field is being changed.
Cite from Volatile vs Static in Java
This means that if two threads update a variable of the same Object concurrently, and the variable is not declared volatile, there could be a case in which one of the thread has in cache an old value.
Given your code, the following can happen:
Thread 1 calls getHelper() and gets null
Thread 2 calls getHelper() and gets null
Thread 1 calls setHelper(42)
Thread 2 calls setHelper(24)
And in this case your trouble starts regarding which Helper object will be used in which thread. The keyword volatile will at least solve the caching problem.
The variable helper is being read by multiple threads simultaneously. At the least, you have to make it volatile or the compiler will begin caching it in registers local to threads and any updates to the variable may not reflect in the main memory. Using volatile, when a thread starts reading a shared variable, it will clear its cache and fetch a fresh value from the global memory. When it finishes reading it, it will flush the contents of its cache into the main memory so that other threads may get the updated value.
I know that in a program that works with multiple threads it's necessary to synchronize the methods because it's possible to have problems like race conditions.
But I cannot understand why we need to synchronize also the methods that need just to read a shared variable.
Look at this example:
public ConcurrentIntegerArray(final int size) {
arr = new int[size];
}
public void set(final int index, final int value) {
lock.lock();
try {
arr[index] = value;
} finally {
lock.unlock();
}
}
public int get(final int index) {
lock.lock();
try {
return arr[index];
} finally {
lock.unlock();
}
}
They did a look on the get and also on the set method. On the set method I understand why. For example if I want to put with Thread1 in index=3 the number 5 and after some milliseconds the Thread2 have to put in index=3 the number 6. Can it happen that I have in index=3 in my array still a 5 instead of a 6 (if I don't do a synchronization on the method set)? This because the Thread1 can have a switch-context and so the Thread2 enter in the same method put the value and after the Thread1 assign the value 5 on the same position So instead of a 6 I have a 5.
But I don't understand why we need (look the example) to synchronize also the method get. I'm asking this question because we need just to read on the memory and not to write.So why we need also on the method get to have a synchronization? Can someone give to me a very simple example?
Both methods need to be synchronized. Without synchronization on the get method, this sequence is possible:
get is called, but the old value isn't returned yet.
Another thread calls set and updates the value.
The first thread that called get now examines the now-returned value and sees what is now an outdated value.
Synchronization would disallow this scenario by guaranteeing that another thread can't just call set and invalidate the get value before it even returns. It would force a thread that calls set to wait for the thread that calls get to finish.
If you do not lock in the get method than a thread might keep a local copy of the array and never refreshes from the main memory. So its possible that a get never sees a value which was updated by a set method. Lock will force the visibility.
Each thread maintain their own copy of value. The synchronized ensures that the coherency is maintained between different threads. Without synchronized, one can never be sure if any one has modified it. Alternatively, one can define the variable as volatile and it will have the same memory effects as synchronized.
The locking action also guarantees memory visibility. From the Lock doc:
All Lock implementations must enforce the same memory synchronization semantics as provided by the built-in monitor lock, [...]:
A successful lock operation has the same memory synchronization effects as a successful Lock action.
A successful unlock operation has the same memory synchronization effects as a successful Unlock action.
Without acquiring the lock, due to memory consistency errors, there's no reason a call to get needs to see the most updated value. Modern processors are very fast, access to DRAM is comparatively very slow, so processors store values they are working on in a local cache. In concurrent programming this means one thread might write to a variable in memory but a subsequent read from a different thread gets a stale value because it is read from the cache.
The locking guarantees that the value is actually read from memory and not from the cache.
Concurrent Hashmap could solve synchronization issue which is seen in hashmap. So adding and removing would be fast if we are using synchronize key work with hashmap. What about checking hashmap size, if mulitple threads checking concurrentHashMap size? do we still need synchronzation key word: something as follows:
public static synchronized getSize(){
return aConcurrentHashmap.size();
}
concurentHashMap.size() will return the size known at the moment of the call, but it might be a stale value when you use that number because another thread has added / removed items in the meantime.
However the whole purpose of ConcurrentMaps is that you don't need to synchronize it as it is a thread safe collection.
You can simply call aConcurrentHashmap.size(). However, you have to bear in mind that by the time you get the answer it might already be obsolete. This would happen if another thread where to concurrently modify the map.
You don't need to use synchronized with ConcurretnHashMap except in very rare occasions where you need to perform multiple operations atomically.
To just get the size, you can call it without synchronization.
To clarify when I would use synchronization with ConcurrentHashMap...
Say you have an expensive object you want to create on demand. You want concurrent reads, but also want to ensure that values are only created once.
public ExpensiveObject get(String key) {
return map.get(key); // can work concurrently.
}
public void put(String key, ExepensiveBuilder builder) {
// cannot use putIfAbsent because it needs the object before checking.
synchronized(map) {
if (!map.containsKey(key))
map.put(key, builder.create());
}
}
Note: This requires that all writes are synchronized, but reads can still be concurrent.
The designers of ConcurrentHashMap thought of giving weightage to individual operations like : get(), put() and remove() over methods which operate over complete HashMap like isEmpty() or size(). This is done because the changes of these methods getting called (in general) are less than the other individual methods.
A synchronization for size() is not needed here. We can get the size by calling concurentHashMap.size() method. This method may return stale values as other thread might modify the map in the meanwhile. But, this is explicitely assumed to be broken as these operations are deprioritized.
ConcorrentHashMap is fail-safe. it won't give any concurrent modification exceptions. it works good for multi threaded operations.
The whole implementation of ConcurrentHashMap is same as HashMap but the while retrieving the elements , HashMap locks whole map restricting doing further modifications which gives concurrent modification exception.'
But in ConcurrentHashMap, the locking happens at bucket level so the chance of giving concurrent modification exception is not present.
So to answer you question here, checking size of ConcurrentHashMap doesn't help because , it keeps chaining based on the operations or modification code that you write on the map. It has size method which is same from the HashMap.
I'm using Collections.synchronizedCollection in Java to protect a Set that I know is getting accessed concurrently by many threads. The Java API warns:
" It is imperative that the user manually synchronize on the returned collection when iterating over it:
Collection c = Collections.synchronizedCollection(myCollection);
...
synchronized(c) {
Iterator i = c.iterator(); // Must be in the synchronized block
while (i.hasNext())
foo(i.next());
}
"
If I use c.contains(obj), is that thread-safe? Internally, obviously, this is iterating over the Collection and seeing if any of the objects in it are equal to obj. My instinct is to assume that this is probably synchronized (it would seem to be a major failing if not), but given previous pains with synchronization, it seems wise to double-check, and a Google search for answers on this hasn't turned up anything.
In itself, a call to contains is safe.
The problem is that one often tests whether a collection contains an element then does something to the collection based on the result.
Most likely, the test and the action should be treated as a single, atomic operation. In that case, a lock on the collection should be obtained, and both operations should be performed in the synchronized block.
Collections.synchronizedCollection() will return a thread safe collection which means
any single method call is thread safe by itself. It depends what you want do. If you want to call couple of methods, java cannot make it thread safe together.
It's safe, because contains itself is synchronized.