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what is the difference between Synchronization and Mutex [duplicate]

Is there any difference between a binary semaphore and mutex or are they essentially the same?

They are NOT the same thing. They are used for different purposes!
While both types of semaphores have a full/empty state and use the same API, their usage is very different.
Mutual Exclusion Semaphores
Mutual Exclusion semaphores are used to protect shared resources (data structure, file, etc..).
A Mutex semaphore is "owned" by the task that takes it. If Task B attempts to semGive a mutex currently held by Task A, Task B's call will return an error and fail.
Mutexes always use the following sequence:
- SemTake
- Critical Section
- SemGive
Here is a simple example:
Thread A Thread B
Take Mutex
access data
... Take Mutex <== Will block
...
Give Mutex access data <== Unblocks
...
Give Mutex
Binary Semaphore
Binary Semaphore address a totally different question:
Task B is pended waiting for something to happen (a sensor being tripped for example).
Sensor Trips and an Interrupt Service Routine runs. It needs to notify a task of the trip.
Task B should run and take appropriate actions for the sensor trip. Then go back to waiting.
Task A Task B
... Take BinSemaphore <== wait for something
Do Something Noteworthy
Give BinSemaphore do something <== unblocks
Note that with a binary semaphore, it is OK for B to take the semaphore and A to give it.
Again, a binary semaphore is NOT protecting a resource from access. The act of Giving and Taking a semaphore are fundamentally decoupled.
It typically makes little sense for the same task to so a give and a take on the same binary semaphore.

A mutex can be released only by the thread that had acquired it.
A binary semaphore can be signaled by any thread (or process).
so semaphores are more suitable for some synchronization problems like producer-consumer.
On Windows, binary semaphores are more like event objects than mutexes.

The Toilet example is an enjoyable analogy:
Mutex:
Is a key to a toilet. One person can
have the key - occupy the toilet - at
the time. When finished, the person
gives (frees) the key to the next
person in the queue.
Officially: "Mutexes are typically
used to serialise access to a section
of re-entrant code that cannot be
executed concurrently by more than one
thread. A mutex object only allows one
thread into a controlled section,
forcing other threads which attempt to
gain access to that section to wait
until the first thread has exited from
that section." Ref: Symbian Developer
Library
(A mutex is really a semaphore with
value 1.)
Semaphore:
Is the number of free identical toilet
keys. Example, say we have four
toilets with identical locks and keys.
The semaphore count - the count of
keys - is set to 4 at beginning (all
four toilets are free), then the count
value is decremented as people are
coming in. If all toilets are full,
ie. there are no free keys left, the
semaphore count is 0. Now, when eq.
one person leaves the toilet,
semaphore is increased to 1 (one free
key), and given to the next person in
the queue.
Officially: "A semaphore restricts the
number of simultaneous users of a
shared resource up to a maximum
number. Threads can request access to
the resource (decrementing the
semaphore), and can signal that they
have finished using the resource
(incrementing the semaphore)." Ref:
Symbian Developer Library

Nice articles on the topic:
MUTEX VS. SEMAPHORES – PART 1: SEMAPHORES
MUTEX VS. SEMAPHORES – PART 2: THE MUTEX
MUTEX VS. SEMAPHORES – PART 3 (FINAL PART): MUTUAL EXCLUSION PROBLEMS
From part 2:
The mutex is similar to the principles
of the binary semaphore with one
significant difference: the principle
of ownership. Ownership is the simple
concept that when a task locks
(acquires) a mutex only it can unlock
(release) it. If a task tries to
unlock a mutex it hasn’t locked (thus
doesn’t own) then an error condition
is encountered and, most importantly,
the mutex is not unlocked. If the
mutual exclusion object doesn't have
ownership then, irrelevant of what it
is called, it is not a mutex.

Since none of the above answer clears the confusion, here is one which cleared my confusion.
Strictly speaking, a mutex is a locking mechanism used to
synchronize access to a resource. Only one task (can be a thread or
process based on OS abstraction) can acquire the mutex. It means there
will be ownership associated with mutex, and only the owner can
release the lock (mutex).
Semaphore is signaling mechanism (“I am done, you can carry on” kind of signal). For example, if you are listening songs (assume it as
one task) on your mobile and at the same time your friend called you,
an interrupt will be triggered upon which an interrupt service routine
(ISR) will signal the call processing task to wakeup.
Source: http://www.geeksforgeeks.org/mutex-vs-semaphore/

Their synchronization semantics are very different:
mutexes allow serialization of access to a given resource i.e. multiple threads wait for a lock, one at a time and as previously said, the thread owns the lock until it is done: only this particular thread can unlock it.
a binary semaphore is a counter with value 0 and 1: a task blocking on it until any task does a sem_post. The semaphore advertises that a resource is available, and it provides the mechanism to wait until it is signaled as being available.
As such one can see a mutex as a token passed from task to tasks and a semaphore as traffic red-light (it signals someone that it can proceed).

At a theoretical level, they are no different semantically. You can implement a mutex using semaphores or vice versa (see here for an example). In practice, the implementations are different and they offer slightly different services.
The practical difference (in terms of the system services surrounding them) is that the implementation of a mutex is aimed at being a more lightweight synchronisation mechanism. In oracle-speak, mutexes are known as latches and semaphores are known as waits.
At the lowest level, they use some sort of atomic test and set mechanism. This reads the current value of a memory location, computes some sort of conditional and writes out a value at that location in a single instruction that cannot be interrupted. This means that you can acquire a mutex and test to see if anyone else had it before you.
A typical mutex implementation has a process or thread executing the test-and-set instruction and evaluating whether anything else had set the mutex. A key point here is that there is no interaction with the scheduler, so we have no idea (and don't care) who has set the lock. Then we either give up our time slice and attempt it again when the task is re-scheduled or execute a spin-lock. A spin lock is an algorithm like:
Count down from 5000:
i. Execute the test-and-set instruction
ii. If the mutex is clear, we have acquired it in the previous instruction
so we can exit the loop
iii. When we get to zero, give up our time slice.
When we have finished executing our protected code (known as a critical section) we just set the mutex value to zero or whatever means 'clear.' If multiple tasks are attempting to acquire the mutex then the next task that happens to be scheduled after the mutex is released will get access to the resource. Typically you would use mutexes to control a synchronised resource where exclusive access is only needed for very short periods of time, normally to make an update to a shared data structure.
A semaphore is a synchronised data structure (typically using a mutex) that has a count and some system call wrappers that interact with the scheduler in a bit more depth than the mutex libraries would. Semaphores are incremented and decremented and used to block tasks until something else is ready. See Producer/Consumer Problem for a simple example of this. Semaphores are initialised to some value - a binary semaphore is just a special case where the semaphore is initialised to 1. Posting to a semaphore has the effect of waking up a waiting process.
A basic semaphore algorithm looks like:
(somewhere in the program startup)
Initialise the semaphore to its start-up value.
Acquiring a semaphore
i. (synchronised) Attempt to decrement the semaphore value
ii. If the value would be less than zero, put the task on the tail of the list of tasks waiting on the semaphore and give up the time slice.
Posting a semaphore
i. (synchronised) Increment the semaphore value
ii. If the value is greater or equal to the amount requested in the post at the front of the queue, take that task off the queue and make it runnable.
iii. Repeat (ii) for all tasks until the posted value is exhausted or there are no more tasks waiting.
In the case of a binary semaphore the main practical difference between the two is the nature of the system services surrounding the actual data structure.
EDIT: As evan has rightly pointed out, spinlocks will slow down a single processor machine. You would only use a spinlock on a multi-processor box because on a single processor the process holding the mutex will never reset it while another task is running. Spinlocks are only useful on multi-processor architectures.

Though mutex & semaphores are used as synchronization primitives ,there is a big difference between them.
In the case of mutex, only the thread that locked or acquired the mutex can unlock it.
In the case of a semaphore, a thread waiting on a semaphore can be signaled by a different thread.
Some operating system supports using mutex & semaphores between process. Typically usage is creating in shared memory.

Mutex: Suppose we have critical section thread T1 wants to access it then it follows below steps.
T1:
Lock
Use Critical Section
Unlock
Binary semaphore: It works based on signaling wait and signal.
wait(s) decrease "s" value by one usually "s" value is initialize with value "1",
signal(s) increases "s" value by one. if "s" value is 1 means no one is using critical section, when value is 0 means critical section is in use.
suppose thread T2 is using critical section then it follows below steps.
T2 :
wait(s)//initially s value is one after calling wait it's value decreased by one i.e 0
Use critical section
signal(s) // now s value is increased and it become 1
Main difference between Mutex and Binary semaphore is in Mutext if thread lock the critical section then it has to unlock critical section no other thread can unlock it, but in case of Binary semaphore if one thread locks critical section using wait(s) function then value of s become "0" and no one can access it until value of "s" become 1 but suppose some other thread calls signal(s) then value of "s" become 1 and it allows other function to use critical section.
hence in Binary semaphore thread doesn't have ownership.

On Windows, there are two differences between mutexes and binary semaphores:
A mutex can only be released by the thread which has ownership, i.e. the thread which previously called the Wait function, (or which took ownership when creating it). A semaphore can be released by any thread.
A thread can call a wait function repeatedly on a mutex without blocking. However, if you call a wait function twice on a binary semaphore without releasing the semaphore in between, the thread will block.

Myth:
Couple of article says that "binary semaphore and mutex are same" or "Semaphore with value 1 is mutex" but the basic difference is Mutex can be released only by thread that had acquired it, while you can signal semaphore from any other thread
Key Points:
•A thread can acquire more than one lock (Mutex).
•A mutex can be locked more than once only if its a recursive mutex, here lock and unlock for mutex should be same
•If a thread which had already locked a mutex, tries to lock the mutex again, it will enter into the waiting list of that mutex, which results in deadlock.
•Binary semaphore and mutex are similar but not same.
•Mutex is costly operation due to protection protocols associated with it.
•Main aim of mutex is achieve atomic access or lock on resource

Mutex are used for " Locking Mechanisms ". one process at a time can use a shared resource
whereas
Semaphores are used for " Signaling Mechanisms "
like "I am done , now can continue"

You obviously use mutex to lock a data in one thread getting accessed by another thread at the same time. Assume that you have just called lock() and in the process of accessing data. This means that you don’t expect any other thread (or another instance of the same thread-code) to access the same data locked by the same mutex. That is, if it is the same thread-code getting executed on a different thread instance, hits the lock, then the lock() should block the control flow there. This applies to a thread that uses a different thread-code, which is also accessing the same data and which is also locked by the same mutex. In this case, you are still in the process of accessing the data and you may take, say, another 15 secs to reach the mutex unlock (so that the other thread that is getting blocked in mutex lock would unblock and would allow the control to access the data). Do you at any cost allow yet another thread to just unlock the same mutex, and in turn, allow the thread that is already waiting (blocking) in the mutex lock to unblock and access the data? Hope you got what I am saying here?
As per, agreed upon universal definition!,
with “mutex” this can’t happen. No other thread can unlock the lock
in your thread
with “binary-semaphore” this can happen. Any other thread can unlock
the lock in your thread
So, if you are very particular about using binary-semaphore instead of mutex, then you should be very careful in “scoping” the locks and unlocks. I mean that every control-flow that hits every lock should hit an unlock call, also there shouldn’t be any “first unlock”, rather it should be always “first lock”.

A Mutex controls access to a single shared resource. It provides operations to acquire() access to that resource and release() it when done.
A Semaphore controls access to a shared pool of resources. It provides operations to Wait() until one of the resources in the pool becomes available, and Signal() when it is given back to the pool.
When number of resources a Semaphore protects is greater than 1, it is called a Counting Semaphore. When it controls one resource, it is called a Boolean Semaphore. A boolean semaphore is equivalent to a mutex.
Thus a Semaphore is a higher level abstraction than Mutex. A Mutex can be implemented using a Semaphore but not the other way around.

Modified question is - What's the difference between A mutex and a "binary" semaphore in "Linux"?
Ans: Following are the differences –
i) Scope – The scope of mutex is within a process address space which has created it and is used for synchronization of threads. Whereas semaphore can be used across process space and hence it can be used for interprocess synchronization.
ii) Mutex is lightweight and faster than semaphore. Futex is even faster.
iii) Mutex can be acquired by same thread successfully multiple times with condition that it should release it same number of times. Other thread trying to acquire will block. Whereas in case of semaphore if same process tries to acquire it again it blocks as it can be acquired only once.

Diff between Binary Semaphore and Mutex:
OWNERSHIP:
Semaphores can be signalled (posted) even from a non current owner. It means you can simply post from any other thread, though you are not the owner.
Semaphore is a public property in process, It can be simply posted by a non owner thread.
Please Mark this difference in BOLD letters, it mean a lot.

Mutex work on blocking critical region, But Semaphore work on count.

http://www.geeksforgeeks.org/archives/9102 discusses in details.
Mutex is locking mechanism used to synchronize access to a resource.
Semaphore is signaling mechanism.
Its up to to programmer if he/she wants to use binary semaphore in place of mutex.

Apart from the fact that mutexes have an owner, the two objects may be optimized for different usage. Mutexes are designed to be held only for a short time; violating this can cause poor performance and unfair scheduling. For example, a running thread may be permitted to acquire a mutex, even though another thread is already blocked on it. Semaphores may provide more fairness, or fairness can be forced using several condition variables.

In windows the difference is as below.
MUTEX: process which successfully executes wait has to execute a signal and vice versa. BINARY SEMAPHORES: Different processes can execute wait or signal operation on a semaphore.

While a binary semaphore may be used as a mutex, a mutex is a more specific use-case, in that only the process that locked the mutex is supposed to unlock it. This ownership constraint makes it possible to provide protection against:
Accidental release
Recursive Deadlock
Task Death Deadlock
These constraints are not always present because they degrade the speed. During the development of your code, you can enable these checks temporarily.
e.g. you can enable Error check attribute in your mutex. Error checking mutexes return EDEADLK if you try to lock the same one twice and EPERM if you unlock a mutex that isn't yours.
pthread_mutex_t mutex;
pthread_mutexattr_t attr;
pthread_mutexattr_init (&attr);
pthread_mutexattr_settype (&attr, PTHREAD_MUTEX_ERRORCHECK_NP);
pthread_mutex_init (&mutex, &attr);
Once initialised we can place these checks in our code like this:
if(pthread_mutex_unlock(&mutex)==EPERM)
printf("Unlock failed:Mutex not owned by this thread\n");

The concept was clear to me after going over above posts. But there were some lingering questions. So, I wrote this small piece of code.
When we try to give a semaphore without taking it, it goes through. But, when you try to give a mutex without taking it, it fails. I tested this on a Windows platform. Enable USE_MUTEX to run the same code using a MUTEX.
#include <stdio.h>
#include <windows.h>
#define xUSE_MUTEX 1
#define MAX_SEM_COUNT 1
DWORD WINAPI Thread_no_1( LPVOID lpParam );
DWORD WINAPI Thread_no_2( LPVOID lpParam );
HANDLE Handle_Of_Thread_1 = 0;
HANDLE Handle_Of_Thread_2 = 0;
int Data_Of_Thread_1 = 1;
int Data_Of_Thread_2 = 2;
HANDLE ghMutex = NULL;
HANDLE ghSemaphore = NULL;
int main(void)
{
#ifdef USE_MUTEX
ghMutex = CreateMutex( NULL, FALSE, NULL);
if (ghMutex == NULL)
{
printf("CreateMutex error: %d\n", GetLastError());
return 1;
}
#else
// Create a semaphore with initial and max counts of MAX_SEM_COUNT
ghSemaphore = CreateSemaphore(NULL,MAX_SEM_COUNT,MAX_SEM_COUNT,NULL);
if (ghSemaphore == NULL)
{
printf("CreateSemaphore error: %d\n", GetLastError());
return 1;
}
#endif
// Create thread 1.
Handle_Of_Thread_1 = CreateThread( NULL, 0,Thread_no_1, &Data_Of_Thread_1, 0, NULL);
if ( Handle_Of_Thread_1 == NULL)
{
printf("Create first thread problem \n");
return 1;
}
/* sleep for 5 seconds **/
Sleep(5 * 1000);
/*Create thread 2 */
Handle_Of_Thread_2 = CreateThread( NULL, 0,Thread_no_2, &Data_Of_Thread_2, 0, NULL);
if ( Handle_Of_Thread_2 == NULL)
{
printf("Create second thread problem \n");
return 1;
}
// Sleep for 20 seconds
Sleep(20 * 1000);
printf("Out of the program \n");
return 0;
}
int my_critical_section_code(HANDLE thread_handle)
{
#ifdef USE_MUTEX
if(thread_handle == Handle_Of_Thread_1)
{
/* get the lock */
WaitForSingleObject(ghMutex, INFINITE);
printf("Thread 1 holding the mutex \n");
}
#else
/* get the semaphore */
if(thread_handle == Handle_Of_Thread_1)
{
WaitForSingleObject(ghSemaphore, INFINITE);
printf("Thread 1 holding semaphore \n");
}
#endif
if(thread_handle == Handle_Of_Thread_1)
{
/* sleep for 10 seconds */
Sleep(10 * 1000);
#ifdef USE_MUTEX
printf("Thread 1 about to release mutex \n");
#else
printf("Thread 1 about to release semaphore \n");
#endif
}
else
{
/* sleep for 3 secconds */
Sleep(3 * 1000);
}
#ifdef USE_MUTEX
/* release the lock*/
if(!ReleaseMutex(ghMutex))
{
printf("Release Mutex error in thread %d: error # %d\n", (thread_handle == Handle_Of_Thread_1 ? 1:2),GetLastError());
}
#else
if (!ReleaseSemaphore(ghSemaphore,1,NULL) )
{
printf("ReleaseSemaphore error in thread %d: error # %d\n",(thread_handle == Handle_Of_Thread_1 ? 1:2), GetLastError());
}
#endif
return 0;
}
DWORD WINAPI Thread_no_1( LPVOID lpParam )
{
my_critical_section_code(Handle_Of_Thread_1);
return 0;
}
DWORD WINAPI Thread_no_2( LPVOID lpParam )
{
my_critical_section_code(Handle_Of_Thread_2);
return 0;
}
The very fact that semaphore lets you signal "it is done using a resource", even though it never owned the resource, makes me think there is a very loose coupling between owning and signaling in the case of semaphores.

Best Solution
The only difference is
1.Mutex -> lock and unlock are under the ownership of a thread that locks the mutex.
2.Semaphore -> No ownership i.e; if one thread calls semwait(s) any other thread can call sempost(s) to remove the lock.

Mutex is used to protect the sensitive code and data, semaphore is used to synchronization.You also can have practical use with protect the sensitive code, but there might be a risk that release the protection by the other thread by operation V.So The main difference between bi-semaphore and mutex is the ownership.For instance by toilet , Mutex is like that one can enter the toilet and lock the door, no one else can enter until the man get out, bi-semaphore is like that one can enter the toilet and lock the door, but someone else could enter by asking the administrator to open the door, it's ridiculous.

I think most of the answers here were confusing especially those saying that mutex can be released only by the process that holds it but semaphore can be signaled by ay process. The above line is kind of vague in terms of semaphore. To understand we should know that there are two kinds of semaphore one is called counting semaphore and the other is called a binary semaphore. In counting semaphore handles access to n number of resources where n can be defined before the use. Each semaphore has a count variable, which keeps the count of the number of resources in use, initially, it is set to n. Each process that wishes to uses a resource performs a wait() operation on the semaphore (thereby decrementing the count). When a process releases a resource, it performs a release() operation (incrementing the count). When the count becomes 0, all the resources are being used. After that, the process waits until the count becomes more than 0. Now here is the catch only the process that holds the resource can increase the count no other process can increase the count only the processes holding a resource can increase the count and the process waiting for the semaphore again checks and when it sees the resource available it decreases the count again. So in terms of binary semaphore, only the process holding the semaphore can increase the count, and count remains zero until it stops using the semaphore and increases the count and other process gets the chance to access the semaphore.
The main difference between binary semaphore and mutex is that semaphore is a signaling mechanism and mutex is a locking mechanism, but binary semaphore seems to function like mutex that creates confusion, but both are different concepts suitable for a different kinds of work.

The answer may depend on the target OS. For example, at least one RTOS implementation I'm familiar with will allow multiple sequential "get" operations against a single OS mutex, so long as they're all from within the same thread context. The multiple gets must be replaced by an equal number of puts before another thread will be allowed to get the mutex. This differs from binary semaphores, for which only a single get is allowed at a time, regardless of thread contexts.
The idea behind this type of mutex is that you protect an object by only allowing a single context to modify the data at a time. Even if the thread gets the mutex and then calls a function that further modifies the object (and gets/puts the protector mutex around its own operations), the operations should still be safe because they're all happening under a single thread.
{
mutexGet(); // Other threads can no longer get the mutex.
// Make changes to the protected object.
// ...
objectModify(); // Also gets/puts the mutex. Only allowed from this thread context.
// Make more changes to the protected object.
// ...
mutexPut(); // Finally allows other threads to get the mutex.
}
Of course, when using this feature, you must be certain that all accesses within a single thread really are safe!
I'm not sure how common this approach is, or whether it applies outside of the systems with which I'm familiar. For an example of this kind of mutex, see the ThreadX RTOS.

Mutexes have ownership, unlike semaphores. Although any thread, within the scope of a mutex, can get an unlocked mutex and lock access to the same critical section of code,only the thread that locked a mutex should unlock it.

As many folks here have mentioned, a mutex is used to protect a critical piece of code (AKA critical section.) You will acquire the mutex (lock), enter critical section, and release mutex (unlock) all in the same thread.
While using a semaphore, you can make a thread wait on a semaphore (say thread A), until another thread (say thread B)completes whatever task, and then sets the Semaphore for thread A to stop the wait, and continue its task.

MUTEX
Until recently, the only sleeping lock in the kernel was the semaphore. Most users of semaphores instantiated a semaphore with a count of one and treated them as a mutual exclusion lock—a sleeping version of the spin-lock. Unfortunately, semaphores are rather generic and do not impose any usage constraints. This makes them useful for managing exclusive access in obscure situations, such as complicated dances between the kernel and userspace. But it also means that simpler locking is harder to do, and the lack of enforced rules makes any sort of automated debugging or constraint enforcement impossible. Seeking a simpler sleeping lock, the kernel developers introduced the mutex.Yes, as you are now accustomed to, that is a confusing name. Let’s clarify.The term “mutex” is a generic name to refer to any sleeping lock that enforces mutual exclusion, such as a semaphore with a usage count of one. In recent Linux kernels, the proper noun “mutex” is now also a specific type of sleeping lock that implements mutual exclusion.That is, a mutex is a mutex.
The simplicity and efficiency of the mutex come from the additional constraints it imposes on its users over and above what the semaphore requires. Unlike a semaphore, which implements the most basic of behaviour in accordance with Dijkstra’s original design, the mutex has a stricter, narrower use case:
n Only one task can hold the mutex at a time. That is, the usage count on a mutex is always one.
Whoever locked a mutex must unlock it. That is, you cannot lock a mutex in one
context and then unlock it in another. This means that the mutex isn’t suitable for more complicated synchronizations between kernel and user-space. Most use cases,
however, cleanly lock and unlock from the same context.
Recursive locks and unlocks are not allowed. That is, you cannot recursively acquire the same mutex, and you cannot unlock an unlocked mutex.
A process cannot exit while holding a mutex.
A mutex cannot be acquired by an interrupt handler or bottom half, even with
mutex_trylock().
A mutex can be managed only via the official API: It must be initialized via the methods described in this section and cannot be copied, hand initialized, or reinitialized.
[1] Linux Kernel Development, Third Edition Robert Love

Mutex and binary semaphore are both of the same usage, but in reality, they are different.
In case of mutex, only the thread which have locked it can unlock it. If any other thread comes to lock it, it will wait.
In case of semaphone, that's not the case. Semaphore is not tied up with a particular thread ID.

Is it possible to allow more number of threads than concurrency level in ConcurrentHashMap?

Suppose we have a ConcurrentHashMap of size 16 .i.e concurrency level 16. If number of threads working on this map is greater than 16 i.e. somewhere 20 to 25 then what will happen to the extra threads whether they will be in waiting state?
If yes then how long they will be in waiting state and internally how they will communicate with other threads?

There's nothing magic about the concurrency level. Even if you have fewer than 16 threads in your example, they still can collide when attempting to access the map.
If the concurrency level is 16, then when any two threads try to put two different keys into the map, there will be one chance in 16 that the two keys are assigned to the same segment. If that's the case, then their access to the segment will have to be serialized. (i.e., one of the threads will have to be blocked until the other thread has finished.)
Of course, if two threads try to access the same key at the same time then it's guaranteed that they will access the same segment and one will be blocked, no matter what the concurrency level.
what will happen to the extra threads whether they will be in waiting state?
Nothing. That's what a synchronized statement does. It does nothing (a.k.a., it "blocks") until the lock becomes available. Then, it acquires the lock, performs the body of the statement, and releases the lock.
how long they will be in waiting state?
A thread that is blocked on a mutex lock will remain blocked until the lock is released by some other thread. In well designed code, that should only be a very short time---just long enough for the other thread to update a few fields.
internally how they will communicate with other threads?
Not sure what you mean by "internally". A thread that is blocked can't communicate with other threads. It can't do anything until it is unblocked.
Maybe you are asking how the operating system knows what to unblock, and when.
When a thread attemts to lock a mutex that already is in use by some other thread, the operating system will suspend the thread, save its state, and put some token that represents the thread into a queue that is associated with the mutex. When the owner of the thread releases the mutex, the OS picks a thread from the queue, transfers ownership of the mutex to the selected thread, restores the thread's saved state, and lets it run.
The algorithms that different operating systems use to select which thread to unblock, and the means of saving and restoring thread contexts (a.k.a., "context switch") are deeper subjects than I have room or time to discuss here.

Is it correct to say a monitor or lock can be owned?

I believe I've seen the expressions "own a monitor", and "own a lock". I'd like to verify that only a monitor can be "owned". And that a lock is "acquired", not owned. If that's wrong, I'd appreciate the correct usage of "own" and "acquire", in the context of Java multithreading.

A lock is kind of data which is logically part of an object’s header on the heap memory. Each object in a JVM has this lock (or mutex) that any program can use to coordinate multi-threaded access to the object. If any thread want to access instance variables of that object; then thread must “own” the object’s lock (set some flag in lock memory area). All other threads that attempt to access the object’s variables have to wait until the owning thread releases the object’s lock (unset the flag).
Once a thread owns a lock, it can request the same lock again multiple times, but then has to release the lock the same number of times before it is made available to other threads. If a thread requests a lock three times, for example, that thread will continue to own the lock until it has “released” it three times.
Monitor is a synchronization construct that allows threads to have both mutual exclusion (using locks) and cooperation i.e. the ability to make threads wait for certain condition to be true (using wait-set).
In other words, along with data that implements a lock, every Java object is logically associated with data that implements a wait-set. Whereas locks help threads to work independently on shared data without interfering with one another, wait-sets help threads to cooperate with one another to work together towards a common goal e.g. all waiting threads will be moved to this wait-set and all will be notified once lock is released. This wait-set helps in building monitors with additional help of lock (mutex).

If a lock is not shared (i.e. can be acquired only once), then whoever holds the single lock currently is the "owner" of the lock.
In the case of a synchronized block, only one Thread is allowed to acquire the lock at the same time (that's the whole purpose here). So that Thread will "own the lock".

Why intrinsic lock like synchronized is discouraged [duplicate]

I'm trying to understand what makes the lock in concurrency so important if one can use synchronized (this). In the dummy code below, I can do either:
synchronized the entire method or synchronize the vulnerable area (synchronized(this){...})
OR lock the vulnerable code area with a ReentrantLock.
Code:
private final ReentrantLock lock = new ReentrantLock();
private static List<Integer> ints;
public Integer getResult(String name) {
.
.
.
lock.lock();
try {
if (ints.size()==3) {
ints=null;
return -9;
}
for (int x=0; x<ints.size(); x++) {
System.out.println("["+name+"] "+x+"/"+ints.size()+". values >>>>"+ints.get(x));
}
} finally {
lock.unlock();
}
return random;
}

A ReentrantLock is unstructured, unlike synchronized constructs -- i.e. you don't need to use a block structure for locking and can even hold a lock across methods. An example:
private ReentrantLock lock;
public void foo() {
...
lock.lock();
...
}
public void bar() {
...
lock.unlock();
...
}
Such flow is impossible to represent via a single monitor in a synchronized construct.
Aside from that, ReentrantLock supports lock polling and interruptible lock waits that support time-out. ReentrantLock also has support for configurable fairness policy, allowing more flexible thread scheduling.
The constructor for this class accepts an optional fairness parameter. When set true, under contention, locks favor granting access to the longest-waiting thread. Otherwise this lock does not guarantee any particular access order. Programs using fair locks accessed by many threads may display lower overall throughput (i.e., are slower; often much slower) than those using the default setting, but have smaller variances in times to obtain locks and guarantee lack of starvation. Note however, that fairness of locks does not guarantee fairness of thread scheduling. Thus, one of many threads using a fair lock may obtain it multiple times in succession while other active threads are not progressing and not currently holding the lock. Also note that the untimed tryLock method does not honor the fairness setting. It will succeed if the lock is available even if other threads are waiting.
ReentrantLock may also be more scalable, performing much better under higher contention. You can read more about this here.
This claim has been contested, however; see the following comment:
In the reentrant lock test, a new lock is created each time, thus there is no exclusive locking and the resulting data is invalid. Also, the IBM link offers no source code for the underlying benchmark so its impossible to characterize whether the test was even conducted correctly.
When should you use ReentrantLocks? According to that developerWorks article...
The answer is pretty simple -- use it when you actually need something it provides that synchronized doesn't, like timed lock waits, interruptible lock waits, non-block-structured locks, multiple condition variables, or lock polling. ReentrantLock also has scalability benefits, and you should use it if you actually have a situation that exhibits high contention, but remember that the vast majority of synchronized blocks hardly ever exhibit any contention, let alone high contention. I would advise developing with synchronization until synchronization has proven to be inadequate, rather than simply assuming "the performance will be better" if you use ReentrantLock. Remember, these are advanced tools for advanced users. (And truly advanced users tend to prefer the simplest tools they can find until they're convinced the simple tools are inadequate.) As always, make it right first, and then worry about whether or not you have to make it faster.
One final aspect that's gonna become more relevant in the near future has to do with Java 15 and Project Loom. In the (new) world of virtual threads, the underlying scheduler would be able to work much better with ReentrantLock than it's able to do with synchronized, that's true at least in the initial Java 15 release but may be optimized later.
In the current Loom implementation, a virtual thread can be pinned in two situations: when there is a native frame on the stack — when Java code calls into native code (JNI) that then calls back into Java — and when inside a synchronized block or method. In those cases, blocking the virtual thread will block the physical thread that carries it. Once the native call completes or the monitor released (the synchronized block/method is exited) the thread is unpinned.
If you have a common I/O operation guarded by a synchronized, replace the monitor with a ReentrantLock to let your application benefit fully from Loom’s scalability boost even before we fix pinning by monitors (or, better yet, use the higher-performance StampedLock if you can).

ReentrantReadWriteLock is a specialized lock whereas synchronized(this) is a general purpose lock. They are similar but not quite the same.
You are right in that you could use synchronized(this) instead of ReentrantReadWriteLock but the opposite is not always true.
If you'd like to better understand what makes ReentrantReadWriteLock special look up some information about producer-consumer thread synchronization.
In general you can remember that whole-method synchronization and general purpose synchronization (using the synchronized keyword) can be used in most applications without thinking too much about the semantics of the synchronization but if you need to squeeze performance out of your code you may need to explore other more fine-grained, or special-purpose synchronization mechanisms.
By the way, using synchronized(this) - and in general locking using a public class instance - can be problematic because it opens up your code to potential dead-locks because somebody else not knowingly might try to lock against your object somewhere else in the program.

From oracle documentation page about ReentrantLock:
A reentrant mutual exclusion Lock with the same basic behaviour and semantics as the implicit monitor lock accessed using synchronized methods and statements, but with extended capabilities.
A ReentrantLock is owned by the thread last successfully locking, but not yet unlocking it. A thread invoking lock will return, successfully acquiring the lock, when the lock is not owned by another thread. The method will return immediately if the current thread already owns the lock.
The constructor for this class accepts an optional fairness parameter. When set true, under contention, locks favor granting access to the longest-waiting thread. Otherwise this lock does not guarantee any particular access order.
ReentrantLock key features as per this article
Ability to lock interruptibly.
Ability to timeout while waiting for lock.
Power to create fair lock.
API to get list of waiting thread for lock.
Flexibility to try for lock without blocking.
You can use ReentrantReadWriteLock.ReadLock, ReentrantReadWriteLock.WriteLock to further acquire control on granular locking on read and write operations.
Have a look at this article by Benjamen on usage of different type of ReentrantLocks

Synchronized locks does not offer any mechanism of waiting queue in which after the execution of one thread any thread running in parallel can acquire the lock. Due to which the thread which is there in the system and running for a longer period of time never gets chance to access the shared resource thus leading to starvation.
Reentrant locks are very much flexible and has a fairness policy in which if a thread is waiting for a longer time and after the completion of the currently executing thread we can make sure that the longer waiting thread gets the chance of accessing the shared resource hereby decreasing the throughput of the system and making it more time consuming.

You can use reentrant locks with a fairness policy or timeout to avoid thread starvation. You can apply a thread fairness policy. it will help avoid a thread waiting forever to get to your resources.
private final ReentrantLock lock = new ReentrantLock(true);
//the param true turns on the fairness policy.
The "fairness policy" picks the next runnable thread to execute. It is based on priority, time since last run, blah blah
also,
Synchronize can block indefinitely if it cant escape the block. Reentrantlock can have timeout set.

One thing to keep in mind is :
The name 'ReentrantLock' gives out a wrong message about other locking mechanism that they are not re-entrant. This is not true. Lock acquired via 'synchronized' is also re-entrant in Java.
Key difference is that 'synchronized' uses intrinsic lock ( one that every Object has ) while Lock API doesn't.

I think the wait/notify/notifyAll methods don't belong on the Object class as it pollutes all objects with methods that are rarely used. They make much more sense on a dedicated Lock class. So from this point of view, perhaps it's better to use a tool that is explicitly designed for the job at hand - ie ReentrantLock.

Lets assume this code is running in a thread:
private static ReentrantLock lock = new ReentrantLock();
void accessResource() {
lock.lock();
if( checkSomeCondition() ) {
accessResource();
}
lock.unlock();
}
Because the thread owns the lock it will allow multiple calls to lock(), so it re-enter the lock. This can be achieved with a reference count so it doesn't has to acquire lock again.

What primitive is used to implement the synchronized keyword?

When we use synchronized keyword in java, which synchronization primitive is used exactly? Lock, Semaphore, Monitor, Mutex ?
EDIT : How JVM implements the lock at the native level ?

At bytecode level, java has monitorenter and monitorexit operations, documented in this page of The Java Virtual Machine Specification, with snippets pasted below (objectref is operand for the operation, taken from stack):
monitorenter snippet
Each object has a monitor associated with it. The thread that executes
monitorenter gains ownership of the monitor associated with objectref. If another thread already owns the monitor associated with objectref, the current thread waits until the object is
unlocked, then tries again to gain ownership. If the current thread
already owns the monitor associated with objectref, it increments a
counter in the monitor indicating the number of times this thread has
entered the monitor. If the monitor associated with objectref is not
owned by any thread, the current thread becomes the owner of the
monitor, setting the entry count of this monitor to 1.
monitorexit snippet
The current thread should be the owner of the monitor associated with
the instance referenced by objectref. The thread decrements the
counter indicating the number of times it has entered this monitor. If
as a result the value of the counter becomes zero, the current thread
releases the monitor. If the monitor associated with objectref
becomes free, other threads that are waiting to acquire that monitor
are allowed to attempt to do so.
So, "monitor" is the answer, and neither this, nor JLS referenced in NPE's answer specify what happens at native code level. If you have a specific platform (CPU and operating system) and a specfic JVM implementation (including version) in mind, you can of course either look at the JVM source (if it is an open source JVM), or ask here.
I also happened across this blog from 1997, which has more details.

From the JLS (§17.1. Synchronization):
The Java programming language provides multiple mechanisms for communicating between threads. The most basic of these methods is synchronization, which is implemented using monitors. Each object in Java is associated with a monitor, which a thread can lock or unlock. Only one thread at a time may hold a lock on a monitor. Any other threads attempting to lock that monitor are blocked until they can obtain a lock on that monitor. A thread t may lock a particular monitor multiple times; each unlock reverses the effect of one lock operation.
Thus "monitor" is the answer to your first question.
As to the second question, this is an unspecified implementation detail.