In a graduate class, we've had to use semaphores to accomplish work with threads.
We were directed to use sem_init along with a bunch of other sem_* procedure but we were not given much information about the details of each of these sem_* methods.
The prototype (and header file) of sem_init is the following:
#include <semaphore.h>
int sem_init(sem_t *sem, int pshared, unsigned int value);
but I don't understand what the pshared value is used for. According to opengroup.org:
If the pshared argument has a non-zero
value, then the semaphore is shared
between processes; in this case, any
process that can access the semaphore
sem can use sem for performing
sem_wait(), sem_trywait(), sem_post(),
and sem_destroy() operations.
but I guess I don't understand the difference between say 1,2, 10, 25, 50000, etc. I think it is saying that if the value is 0 then the semaphore is not shared. (But then, what is the point?)
How do I appropriately use this pshared parameter?
The GLIBC version of sem_init (what you get if you man sem_init on Linux) has this to say:
"The pshared argument indicates whether this semaphore is to be
shared between the threads of a process, or between processes."
So pshared is a boolean value: in practice meaningful values passed to it are false (0) and true (1), though any non-0 value will be treated as true. If you pass it 0 you will get a semaphore that can be accessed by other threads in the same process -- essentially an in-process lock. You can use this as a mutex, or you can use it more generally for the resource-counting properties of a semaphore. Arguably if pthreads supported a semaphore API you wouldn't need this feature of sem_init, but semaphores in Unix precede pthreads by quite a bit of time.
It would be better if the boolean was some kind of enumeration (e.g. SEM_PROCESS_PRIVATE vs SEM_PROCESS_SHARED), because then you wouldn't have had this question, but POSIX semaphores are a fairly old API as these things go.
I would say that there is no significant difference between the value s 1, 2, 5 and so on with respect to the shared parameter. Probably it is written that way because when the API was first created, C did not have boolean types.
The pshared argument indicates whether this semaphore is to be shared between the threads of a process, or between processes.
If pshared has the value 0, then the semaphore is shared between the threads of a process, and should be located at some address that is visible to all threads (e.g., a global variable, or a variable allocated dynamically on the heap).
If pshared is nonzero, then the semaphore is shared between processes, and should be located in a region of shared memory (see shm_open(3), mmap(2), and shmget(2)). (Since a child created by fork(2) inherits its parent's memory mappings, it can also access the semaphore.) Any process that can access the shared memory region can operate on the semaphore using sem_post(3), sem_wait(3), etc.
The pshared argument indicates whether this semaphore is to be shared between the threads of a process, or between processes.If pshared has the value 0, then the semaphore is shared between the threads of a process, and should be located at some address that is visible to all threads.If pshared is nonzero, then the semaphore is shared betweenprocesses, and should be located in a region of shared memory.
Related
In an article about c++11 memory order, author show an example reasoning "threads lib will not work in c++03"
for (...){
...
if (mt) pthread_mutex_lock(...);
x=...x...
if (mt) pthread_mutex_unlock(...);
}
//should not have data-race
//but if "clever" compiler use a technique called
//"register promotion" , code become like this:
r = x;
for (...){
...
if (mt) {
x=r; pthread_mutex_lock(...); r=x;
}
r=...r...
if (mt) {
x=r; pthread_mutex_unlock(...); r=x;
}
x=r;
There are 3 question:
1.Is this promotion only break the mutex protection in c++03?What about c language?
2.c++03 thread libs become unwork?
3.Any other promotion may caused same problem?
If it's wrong example, then thread libs work, what about the 《Threads Cannot be Implemented as a Library》by Hans Boehm.
POSIX functions pthread_mutex_lock and pthread_mutex_unlock are memory barriers, the compiler and/or CPU cannot reorder loads and stores around them. Otherwise the mutexes would be useless. That article is probably inaccurate.
See POSIX 4.12 Memory Synchronization:
Applications shall ensure that access to any memory location by more than one thread of control (threads or processes) is restricted such that no thread of control can read or modify a memory location while another thread of control may be modifying it. Such access is restricted using functions that synchronize thread execution and also synchronize memory with respect to other threads. The following functions synchronize memory with respect to other threads: [see the list on the website]
For single thread code, the state in the abstract machine is not directly observable: objects that aren't volatile are not guaranteed to have any particular state when you pause the only thread with a signal and observe it via ptrace or the equivalent. The only requirement is that the program execution has the same observable behavior as a behavior of one possible execution of the abstract machine.
The observables are the interactions with external world; basically, input/output on streams and actions on volatile objects.
A compiler for mono-thread code can generate code that perform operations on global variables or other object that happen to be shared between threads, as long as the single thread semantic is respected. This is obviously the case if a global variable to changed in such a way that it gets back its original value.
For example, a compiler might emit code that increment then decrement a variable, at least in some rare cases; the goal would be to emit simple code, at the cost of the occasional few unneeded operations.
Such changes to shared variables that don't exist in the abstract machine would obviously break multithreaded code that concurrently performs a real operation; such code does not have any race condition on the accesses of the shared variable, that are properly serialized, but the generated code introduced a race that breaks the program.
Does a mutex lock access to variables globally, or just those in the same scope as the locked mutex?
Note that I had to change the title of this question, as a lot of answers seem to be confused as to what I was asking. This is not a question about the scope (global or otherwise) of a "mutex object", it is a question about what scope of variables are "locked" by a mutex.
I believe the answer to be that a mutex locks access to all variables, ie; all global and locally scoped variables. (This is a result of a mutex blocking thread execution rather than access to specific regions of memory.)
I am attempting to understand Mutexes.
I was attempting to understand what sections of memory, or equivalently, which variables, a mutex would lock.
However my understanding from reading around online is that Mutexes do not lock memory, they lock (or block) simultaneously running threads which are all members of the same process. (Is that correct?)
https://mortoray.com/2011/12/16/how-does-a-mutex-work-what-does-it-cost/
So my question has become simply "are mutexes global?"
... or are they perhaps "generally speaking global, but the stackoverflow community can imagine some special cases in which they are not?"
When originally considering my question, I was interested in things such as those shown in the following example.
// both in global scope, this mutex will lock any global scope variable?
int global_variable;
mutex global_variable_mutex;
int main()
{
// one thread operates here and locks global_variable_mutex
// before reading/writing
{
// local variables in a loop
// launch some threads here, and wait later
int local_variable;
mutex local_variable_mutex;
// wait for launched thread to return
// does the mutex here prevent data races to the variable
// global_variable ???
}
}
One may assume this is pseudo-code for C++ or C, or any other similarly relevant language.
2021 edit: Question title has been changed to better reflect the contents of the question and associated answers.
So my question has become simply "are mutexes global?"
No. A mutex has a lock() and an unlock() method, and the only thing a mutex does is cause its lock() call (from any thread) not to return for as long as another thread has that mutex locked. When the thread that was holding the mutex locked calls unlock(), that is when the lock() call will return in the first thread. That way it is guaranteed that only a single thread will be holding the mutex-lock (i.e. executing in the region between its lock() call and its unlock() call) at any given time.
That's really all there is to it. So a mutex will effect only the threads that call lock() on that particular mutex, and nothing else.
Mutex stands for "Mutual Exclusion" - using one correctly ensures that only one thread at a time will ever be executing any "critical section" protected by the same mutex.
If there are some variables you only ever modify inside critical sections protected by the same mutex, your code doesn't have a data race. No matter whether they're global, static, or pointed to by different variables in different threads or any other way two threads might have a reference to the same object.
When I asked this question I was confused...
When I originally asked this question, I was confused because I had no conceputal understanding of how a "mutex" functions in hardware, whereas I did have a conceptual understanding of many other things that exist in hardware. (For example, how a compiler converts text into machine readable instructions. How cache and memory work. How graphics or coprocessors work. How network hardware and interfaces work, etc.)
Misconception 1: Mutex does not lock memory locations
When I first heard about Mutex, long before writing this question, I misunderstood a mutex to be a feature which locks regions of memory. (That region might be global.)
This is not what happens. Other threads and processes can continue to access main memory and cache if another thread locks a mutex. You can see immediatly why such a design would be inefficient, since it would block all other system processes, for the sake of synchronizing one.
Misconception 2: The scope in which a mutex object is declared is irrelevant
The context of this is C code, and C like languages where you have scoped blocks defined by { and } however the same logic could apply to Python where scope is defined by indentation.
I believe that this misunderstanding came from the existance of scoped_lock objects, and similar concepts where scope is used to manage the lifetime (locking and unlocking, resources) of a Mutex object.
One could also argue that since pointers and references to a Mutex can be passed around a program, the scope of a Mutex couldn't be used to define what variables are "locked" by a mutex.
For example, I misunderstood the following snippet:
{
int x, y, z;
Mutex m;
m.lock();
}
I believed that the above snippet would lock access to variables x, y and z from all other threads because x, y and z are declared in the same scope as the mutex m. This is also not how a mutex works.
Understanding 1: Mutex is typically implemented in hardware using atomic operations
Atomic operations are completely seperate from the concept of mutex, however they are a prerequisite to understanding how a mutex can exist, and how it can work.
When a CPU executes something like c = a + b, this involves a sequence of individual (atomic) operations. The word Atom is derived from Atomos meaning "indivisible", or "fundamental". (Atoms are divisible, but when theorists of Ancient Greece originally concieved of the objects from which matter was composed, they assumed that particles must be divisible down to some fundamental smallest possible component, which itself is indivisible. They were not too far wrong, since an atom is made from other fundamental particles which so far we understand to be indivisible.)
Returning to the point: c = a + b is something like the following:
load a from memory into register 1
load b from memory into register 2
do operation add: add contents of register 2 to register 1, result is in register 1
save register 1 to memory c
The add operation might take several clock cycles, and loading/saving to memory takes typically of order 100 clock cycles on modern x86 machines. However each operation is atomic in the sense that a single CPU instruction is being completed, and this instruction cannot be divided into any smaller step of smaller instructions. The instructions are themselves fundamental computing operations.
With that understood, there exists a set of atomic instructions which can do things such as:
load a value from memory increment it and save it to memory
load a value from memory decrement it and save it to memory
load a value from memory, compare it to a value which is already loaded into a register, and branch depending on the comparison result
Note that such operations are typically significantly slower than their non-atomic sequence counterparts. This is because optimizations such as pipelining are forfit when executing the above instructions. (I think?)
At this point my knowledge becomes a bit less accurate and more hand-wavey, but as far as I understand, these operations are typically implemented by having some digital logic inside the processor which blocks all other processes from running while these atomic operations (listed above) are executing.
Meaning: If there are 8 CPU cores running, if one core encounters an instruction like the above, it signals the other cores to stop running until it has finished that atomic operation. (It is at least something approximatly along these lines.)
Understanding 2: Actual mutex operation
Given the above, it is possible to implement a mutex using these atomic machine instructions. Other answers posted here suggest possible ways of doing it including something similar to reference counting. (Semaphore.)
How an acutal mutex in C++ works is this:
Each mutex object has a variable in memory associated with it, the value of this variable indicates whether a mutex is locked or not
This mutex variable is updated using the special atomic operations that a CPU supports for the purpose of allowing a mutex to be programmed
Elsewhere in memory there are some other variables/data which you want to protect/synchronize access to
This synchronization is done using the mutex variable/data
Before a thread reads/writes to some data/variable which needs to be accessed mutually exclusively by all threads which operate on it, that thread must first "lock" the special mutex data/variable
This is done using the atomic operations built into a CPU for the purpose of supporting mutex programming
So you see, the data which is "locked" and accessed mutually exclusively is entirely independent from the actual data used to store the state of the mutex.
If another thread wants to read/write the data which must be accessed mutually exclusively, it will try to lock the mutex. If the mutex is already locked, that means another thread has the right to access this data, and no other thread is permitted to, therefore this thread will typically go to sleep, and will be re-woken by the operating system when the mutex is next unlocked.
It is important to note the operating system thread (kernel) is critically involved in the mutex process. Typically, before a thread sleeps, it will tell the operating sytem that it wishes to be woken up again when the mutex is free. The operating system is also notified when other threads lock or unlock a mutex. Hence synchronization of information about the state of a mutex is passed via messages through the operating system kernel.
This is why writing a multiple thread OS kernel is (proabably) impossible (if not very difficult). I don't know if this has actually been done successfully. It sounds like a difficult problem which might be the subject of current CS research.
This is pretty much everything I know about the subject. Obviously my knowledge is not absolute...
Note: Feel free to correct my Greek history or x86 Machine Instruction knowledge in the comments section. No doubt not everything here is perfectly accurate.
As your question suggests, I assume you are asking your question independent of any programming language.
First it is important to understand what is a mutex and how it works? A mutex is a binary semaphore. Then what is a semaphore? A semaphore is an integer with following attributes,
You can initialize it into any permitted value (For a mutex, it is 1 or 0).
A thread can access the semaphore and it can increment or decrement its integer value.
When a thread decrements it,
If the result is positive or zero, that thread can continue its process.
If the result is negative, that thread will be waiting and the semaphore value will not be further decremented by any later thread.
If a thread increments it, (in that case semaphore value will be either positive or 0) and the result is 0, one of the waiting threads can continue execution.
So when there's a situation where a thread is trying to access a shared resource it will decrement the mutex value (from 0, so that other thread is waiting). And when it finishes, it will increment the mutex value (So that the waiting thread can continue). That's how the access control happens by means of a mutex (Binary semaphore).
I think you understand that your question is a non-applicable one here. As a simple answer for
So my question has become simply "are mutexes global?"
is simply NO.
A mutex has whatever scope you assign to it. It can be global or local again based on where and how you declare it. If for example you declare a mutex in global memory in a place where you can access it globally, then it is indeed global. If instead you declare it at function or private class scope level, then only that function or class will have access to it.
That said, in order to be useful for synchronization, the mutex needs to be declared in a scope that can be accessed by the threads needing to synchronize on it. Whether that's at global scope or some local scope depends on your program structure. I'd advise declaring it at the highest scope accessible to the threads but no higher.
In your particular example, the mutex is indeed global because you've declared it in global memory.
Locking doesn't operate on the variables it protects, it just works by giving threads a way to arrange that only one thread at a time will be doing something (like reading+writing a data structure). And that it will be finished, with memory effects visible, before the next thread's turn to read and maybe modify that data. (A readers+writers lock allows multiple readers but only one writer).
Any thread that can access the mutex object can lock / unlock it. The mutex object itself is a normal variable that you can put in any scope you want, even a local variable and then put a pointer to it somewhere that other threads can see. (Although normally you wouldn't do that.)
Mutex is named for "Mutual Exclusion" - using one correctly ensures that only one thread at a time will ever be executing any "critical section" (wikipedia) protected by the same mutex. Separate mutexes can allow different threads to hold different locks. Different functions or blocks that use the same mutex (normally because they access the same data) won't both run at once.
If there are some variables you only ever modify inside critical sections protected by the same mutex, those accesses won't be data race, and if you don't have other bugs, your code is thread-safe. No matter whether they're global, static, or pointed to by different variables in different threads or any other way two threads might have a reference to the same object.
If you write code that accesses shared data without taking a lock on a mutex, it might see a partially-updated value, especially for a struct with multiple pointers / integers. (And in C++, simultaneous accesses to non-atomic variables is undefined behaviour if they're not all reads).
Locking is a cooperative activity, normally nothing stops you from getting it wrong. If you're familiar with file locking, you may have heard of advisory vs. mandatory locks (the OS will deny open calls by other programs). Mutexes in multi-threaded programs are advisory; no memory protection or other hardware mechanism stops another thread from executing code that accesses the bytes of an object.
(At a low enough level, that's actually useful for lock-free atomics, especially with some control over ordering of those operations from memory barriers and/or release-store / acquire-load. And CPU cache hardware is up to the task of maintaining coherency from multiple accesses. But if you use locking, you don't have to worry about any of that. If you use locking incorrectly, understanding the possible symptoms might help identify that there is a locking problem.)
Some programs have phases where only a single thread is running, or only one that would need to touch certain variables, so enforced locking for every access to a variable isn't something that every language provides. (C++ std::atomic<T> is sort of like that; every access is as-if there was a lock/unlock of a lock protecting just that T object, except it's limited to operations that most CPUs can do without needing to lock/unlock a separate lock. Unless you use a large T, then there actually is a lock. Or if you use a memory order weaker than the default seq_cst, you can see orderings that wouldn't have been possible if all accesses acquiring/releasing locks.)
Besides, consistency between multiple variables is often important, so it matters that you hold one lock across multiple operations on multiple variables, or multiple members of the same struct.
Some tools can help detect code that doesn't respect a mutex while other threads are running, though, like clang -fsanitize=thread.
I read one of the differences between semaphore and mutex is in case of mutex the process/thread (which ever is having the lock) can only release the lock. But in the case of the semaphore any other process can release the semaphore. My doubt arises when a process that does not have the semaphore with it can release the semaphore. What is the use of having a semaphore?
Let's say I have two processes A and B. Assume process A is having a semaphore with it and executing some critical task. Now let us say process B sends a signal to release the semaphore. In this scenario, will process A release the semaphore even if it is executing some critical task?
You are making half-sense. It is not about ownership. Partner-release in semaphores (and mutexes) is usable, for instance, in my favorite interview question of thread ping-pong. As a matter of fact, I have specifically tried to partner-release a mutex on 3 implementations available to me at a time (Linux/Solaris/AIX) and partner-release did work for mutexes as expected - i.e. mutex was successsfully released and threads blocking on it resumed execution. However, this is, of course, prohibited by Posix.
I think you might be confused on the whole set of differences between a semaphore and a mutex. A mutex provides mutual exclusion. A semaphore counts until it reaches a level where it starts excluding. A semaphore that counted to one would give similar semantics to a mutex though.
A good example would be a television set. Only so many people can watch the same television set, so protecting it with a semaphore would make sense. Anyone can stop watching the television. The remote control for the television can only be operated by one person at a time though, so you could protect it with a mutex.
Some reading...
https://en.wikipedia.org/wiki/Mutual_exclusion
https://en.wikipedia.org/wiki/Semaphore_%28programming%29
"Let's say I have two processes A and B. Assume process A is having a semaphore with it and executing some critical task. Now let us say process B sends a signal to release the semaphore. In this scenario, will process A release the semaphore even if it is executing some critical task?"
One key point to note here is the role of OS kernel. Process B can't send a signal to Process A 'to release the semaphore'. What it can do is request the kernel to give it access to the resource. Process A had requested the kernel and the kernel granted it access to the resource.
Now process A, after it finishes its job, will let the kernel know that it is done with the resource and then kernel grants access to B.
"My doubt arises when a process that does not have the semaphore with it can release the semaphore. What is the use of having a semaphore?"
The key difference between a mutex and a semaphore is, a semaphore serializes access to multiple instances of a resource. Mutex does the same when there is one instance of the resource.
A count is maintained by kernel in case of semaphore and mutex is a special case where the count is 1.
Consider the processes as customers waiting in line at a bank.
The use of semaphore is analogous to the case where there are multiple tellers serving the customers. Usage of mutex is analogous to the case where there is just one teller.
Say there are processes A, B and C that need concurrent access to a resource (lock, file or a data structure in memory, etc.). Further suppose there are 2 instances of the resource. So at most two processes can be granted access at a time.
Process A requests access to an instance of the resource following the required semantics. This request to the kernel involves data structures to identify the resource and maximum number of instances as 2. kernel creates the semaphore with a count of 2, grants A access to the resource and decrements the count to 1, because now only one other process can get access.
Now process B requests access to the resource by following the same semantics. Kernel grants it access and decrements the count to 0.
Now process C requests access, but kernel keeps it in waiting state, because count is 0 and no more than 2 processes can get concurrent access.
Process A is done with the resource and lets kernel know. Kernel notices this and grants access to process C that has been waiting.
In case of mutex, kernel grants access to the resource only one process at a time.
A normal binary semaphore is basically used for synchronization. However, the mutex is for exclusive access to a resource. A mutex is a special variant of semaphore that allows only one locker at a time and with more stringency on ownership than a normal semaphore such as the mutex should be released only by the thread that acquired it. Also, please note that in case of pthreads, fast mutex may not check for this error related to ownership, whereas the error checking mutex shall return error.
For the query related to 2 process A and B, the Process A shall intimate via kernel that it is done with its critical work so that the resource can be made available for waiting processes like B.
You could find some related information in this link too :
When should we use mutex and when should we use semaphore
There is no such thing as "having" a semaphore. Semaphores don't have ownership like mutexes do. The code you describe would simply be buggy. Mutexes won't work if your code is buggy either.
Consider the most classic example of a semaphore -- allowing one train at a time on a section of track. You could implement this with a mutex if the train is a thread. The train would lock the track mutex before going on the track and unlock it after leaving the track.
But what if the train itself is multi-threaded? Which thread should own the track?
And what if the signalling devices are the threads, not the train? Here, the signalling device that detects the train entering the track has to lock the track while the signalling device that detects the train leaving the track has to unlock it.
Mutexes are suitable for cases where there is something that is owned by a particular thread for a short period of time. That thread can "own" the mutex. Semaphores are useful for cases where there is no thread to own anything or nothing for the thread to own.
For a class I'm taking I've been doing some work directly with the clone() system call in Linux. I got curious about how it actually worked and started doing some digging. What is confusing me is that it seems to rely on some of the same underpinnings as fork() functionality (they call the same do_fork() function albeit with different arguments). On one hand, this makes sense to me as a thread is really just a light-weight process but I was always under the impression that there were some significant differences between the way a thread was created an the way a process was created. I did some digging into the implementation of do_fork() and subsequently copy_process() (which do_fork() calls) but I haven't been able to convince myself I understand what's going on.
So, to the guru's out there, am I missing something or is this actually how it works? Are there flags that basically tell the OS just how much to copy as well as what instruction to begin execution of the new task at (I'm thinking the answer has to be yes, but I'm just not sure how they translate)?
Below is the code I'm looking at, perhaps you could explain how the arguments that are passed in control whether a light-weight or heavy-weight process is created.
asmlinkage int sys_fork(struct pt_regs *regs){
#ifdef CONFIG_MMU
return do_fork(SIGCHLD, regs->ARM_sp, regs, 0, NULL, NULL);
#else
/* can not support in nommu mode */
return(-EINVAL);
#endif
}
asmlinkage int sys_clone(unsigned long clone_flags, unsigned long newsp,
int __user *parent_tidptr, int tls_val,
int __user *child_tidptr, struct pt_regs *regs)
{
if (!newsp)
newsp = regs->ARM_sp;
return do_fork(clone_flags, newsp, regs, 0, parent_tidptr, child_tidptr);
}
Thanks!
Actually, at the conceptual level, the Linux kernel doesn't know anything about processes or threads, it only knows about "tasks".
A Linux task can be a process, a thread or something in between. (Incidentally, this means that the strange children that vfork() creates fit perfectly well into the Linux "task" paradigm).
Now, tasks can share several things, see all the CLONE_* flags in the manpage for clone(2). (Not all these flags can be described as sharing, some specify more complex behaviours).
Or new tasks can choose to have their own copies of the respective resources. And since 2.6.16, they can do so after having been started, see unshare(2).
For instance, the only difference between a vfork() and a fork() call, is that vfork() has CLONE_VM and CLONE_VFORK set. CLONE_VM makes it share its parent's memory (the same way threads share memory), while CLONE_VFORK makes the parent block until the child releases its memory mappings (by calling execve() or _exit()).
Note that Linux is not the only OS to generalize processes and threads in this manner. Plan 9 has rfork().
Nothing in the clone manpage suggests that it's "lightweight".
The critical difference is that fork creates a new address space, while clone optionally shares the address space between the parent and child, as well as file handles and so forth.
This shared address space enables lightweight IPC later on, but the process itself is not slimmer.
I understand that the difference between all the three clone,fork and vfork is in the flags because finally all the three calls the do_fork() in kernel
fork()-->C_lib-->sys_fork()-->do_fork()
vfork()-->C_lib-->sys_vfork()-->do_fork()
clone()-->C_lib-->sys_clone()-->do_fork()
The difference between the fork and vfork is that vfork guarantees that child will execute first and parent will block until child calls exit or exec. vfork passes extra flag that is CLONE_VM, this flag ask kernel not to duplicate the page table, the reason is simple the child will either do exit or exec, if child exits nothing would be done, if child does the exec the page table will definitely be changed. i hope the fork and vfork flags are clear now at kernel level.
Now lets look at the clone flags
The main usage of clone is to implement thread, where the memory space shared other then stack. Along with same parameter as fork and vfork the clone also takes the function pointer as parameter, which is called as soon as the child process is created.
When should we use mutex and when should we use semaphore ?
Here is how I remember when to use what -
Semaphore:
Use a semaphore when you (thread) want to sleep till some other thread tells you to wake up. Semaphore 'down' happens in one thread (producer) and semaphore 'up' (for same semaphore) happens in another thread (consumer)
e.g.: In producer-consumer problem, producer wants to sleep till at least one buffer slot is empty - only the consumer thread can tell when a buffer slot is empty.
Mutex:
Use a mutex when you (thread) want to execute code that should not be executed by any other thread at the same time. Mutex 'down' happens in one thread and mutex 'up' must happen in the same thread later on.
e.g.: If you are deleting a node from a global linked list, you do not want another thread to muck around with pointers while you are deleting the node. When you acquire a mutex and are busy deleting a node, if another thread tries to acquire the same mutex, it will be put to sleep till you release the mutex.
Spinlock:
Use a spinlock when you really want to use a mutex but your thread is not allowed to sleep.
e.g.: An interrupt handler within OS kernel must never sleep. If it does the system will freeze / crash. If you need to insert a node to globally shared linked list from the interrupt handler, acquire a spinlock - insert node - release spinlock.
A mutex is a mutual exclusion object, similar to a semaphore but that only allows one locker at a time and whose ownership restrictions may be more stringent than a semaphore.
It can be thought of as equivalent to a normal counting semaphore (with a count of one) and the requirement that it can only be released by the same thread that locked it(a).
A semaphore, on the other hand, has an arbitrary count and can be locked by that many lockers concurrently. And it may not have a requirement that it be released by the same thread that claimed it (but, if not, you have to carefully track who currently has responsibility for it, much like allocated memory).
So, if you have a number of instances of a resource (say three tape drives), you could use a semaphore with a count of 3. Note that this doesn't tell you which of those tape drives you have, just that you have a certain number.
Also with semaphores, it's possible for a single locker to lock multiple instances of a resource, such as for a tape-to-tape copy. If you have one resource (say a memory location that you don't want to corrupt), a mutex is more suitable.
Equivalent operations are:
Counting semaphore Mutual exclusion semaphore
-------------------------- --------------------------
Claim/decrease (P) Lock
Release/increase (V) Unlock
Aside: in case you've ever wondered at the bizarre letters (P and V) used for claiming and releasing semaphores, it's because the inventor was Dutch. In that language:
Probeer te verlagen: means to try to lower;
Verhogen: means to increase.
(a) ... or it can be thought of as something totally distinct from a semaphore, which may be safer given their almost-always-different uses.
It is very important to understand that a mutex is not a semaphore with count 1!
This is the reason there are things like binary semaphores (which are really semaphores with count 1).
The difference between a Mutex and a Binary-Semaphore is the principle of ownership:
A mutex is acquired by a task and therefore must also be released by the same task.
This makes it possible to fix several problems with binary semaphores (Accidental release, recursive deadlock, and priority inversion).
Caveat: I wrote "makes it possible", if and how these problems are fixed is up to the OS implementation.
Because the mutex has to be released by the same task it is not very good for the synchronization of tasks. But if combined with condition variables you get very powerful building blocks for building all kinds of IPC primitives.
So my recommendation is: if you got cleanly implemented mutexes and condition variables (like with POSIX pthreads) use these.
Use semaphores only if they fit exactly to the problem you are trying to solve, don't try to build other primitives (e.g. rw-locks out of semaphores, use mutexes and condition variables for these)
There is a lot of misunderstanding between mutexes and semaphores. The best explanation I found so far is in this 3-Part article:
Mutex vs. Semaphores – Part 1: Semaphores
Mutex vs. Semaphores – Part 2: The Mutex
Mutex vs. Semaphores – Part 3 (final part): Mutual Exclusion Problems
While #opaxdiablo answer is totally correct I would like to point out that the usage scenario of both things is quite different. The mutex is used for protecting parts of code from running concurrently, semaphores are used for one thread to signal another thread to run.
/* Task 1 */
pthread_mutex_lock(mutex_thing);
// Safely use shared resource
pthread_mutex_unlock(mutex_thing);
/* Task 2 */
pthread_mutex_lock(mutex_thing);
// Safely use shared resource
pthread_mutex_unlock(mutex_thing); // unlock mutex
The semaphore scenario is different:
/* Task 1 - Producer */
sema_post(&sem); // Send the signal
/* Task 2 - Consumer */
sema_wait(&sem); // Wait for signal
See http://www.netrino.com/node/202 for further explanations
See "The Toilet Example" - http://pheatt.emporia.edu/courses/2010/cs557f10/hand07/Mutex%20vs_%20Semaphore.htm:
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
Mutex is to protect the shared resource.
Semaphore is to dispatch the threads.
Mutex:
Imagine that there are some tickets to sell. We can simulate a case where many people buy the tickets at the same time: each person is a thread to buy tickets. Obviously we need to use the mutex to protect the tickets because it is the shared resource.
Semaphore:
Imagine that we need to do a calculation as below:
c = a + b;
Also, we need a function geta() to calculate a, a function getb() to calculate b and a function getc() to do the calculation c = a + b.
Obviously, we can't do the c = a + b unless geta() and getb() have been finished.
If the three functions are three threads, we need to dispatch the three threads.
int a, b, c;
void geta()
{
a = calculatea();
semaphore_increase();
}
void getb()
{
b = calculateb();
semaphore_increase();
}
void getc()
{
semaphore_decrease();
semaphore_decrease();
c = a + b;
}
t1 = thread_create(geta);
t2 = thread_create(getb);
t3 = thread_create(getc);
thread_join(t3);
With the help of the semaphore, the code above can make sure that t3 won't do its job untill t1 and t2 have done their jobs.
In a word, semaphore is to make threads execute as a logicial order whereas mutex is to protect shared resource.
So they are NOT the same thing even if some people always say that mutex is a special semaphore with the initial value 1. You can say like this too but please notice that they are used in different cases. Don't replace one by the other even if you can do that.
Trying not to sound zany, but can't help myself.
Your question should be what is the difference between mutex and semaphores ?
And to be more precise question should be, 'what is the relationship between mutex and semaphores ?'
(I would have added that question but I'm hundred % sure some overzealous moderator would close it as duplicate without understanding difference between difference and relationship.)
In object terminology we can observe that :
observation.1 Semaphore contains mutex
observation.2 Mutex is not semaphore and semaphore is not mutex.
There are some semaphores that will act as if they are mutex, called binary semaphores, but they are freaking NOT mutex.
There is a special ingredient called Signalling (posix uses condition_variable for that name), required to make a Semaphore out of mutex.
Think of it as a notification-source. If two or more threads are subscribed to same notification-source, then it is possible to send them message to either ONE or to ALL, to wakeup.
There could be one or more counters associated with semaphores, which are guarded by mutex. The simple most scenario for semaphore, there is a single counter which can be either 0 or 1.
This is where confusion pours in like monsoon rain.
A semaphore with a counter that can be 0 or 1 is NOT mutex.
Mutex has two states (0,1) and one ownership(task).
Semaphore has a mutex, some counters and a condition variable.
Now, use your imagination, and every combination of usage of counter and when to signal can make one kind-of-Semaphore.
Single counter with value 0 or 1 and signaling when value goes to 1 AND then unlocks one of the guy waiting on the signal == Binary semaphore
Single counter with value 0 to N and signaling when value goes to less than N, and locks/waits when values is N == Counting semaphore
Single counter with value 0 to N and signaling when value goes to N, and locks/waits when values is less than N == Barrier semaphore (well if they dont call it, then they should.)
Now to your question, when to use what. (OR rather correct question version.3 when to use mutex and when to use binary-semaphore, since there is no comparison to non-binary-semaphore.)
Use mutex when
1. you want a customized behavior, that is not provided by binary semaphore, such are spin-lock or fast-lock or recursive-locks.
You can usually customize mutexes with attributes, but customizing semaphore is nothing but writing new semaphore.
2. you want lightweight OR faster primitive
Use semaphores, when what you want is exactly provided by it.
If you dont understand what is being provided by your implementation of binary-semaphore, then IMHO, use mutex.
And lastly read a book rather than relying just on SO.
I think the question should be the difference between mutex and binary semaphore.
Mutex = It is a ownership lock mechanism, only the thread who acquire the lock can release the lock.
binary Semaphore = It is more of a signal mechanism, any other higher priority thread if want can signal and take the lock.
All the above answers are of good quality,but this one's just to memorize.The name Mutex is derived from Mutually Exclusive hence you are motivated to think of a mutex lock as Mutual Exclusion between two as in only one at a time,and if I possessed it you can have it only after I release it.On the other hand such case doesn't exist for Semaphore is just like a traffic signal(which the word Semaphore also means).
As was pointed out, a semaphore with a count of one is the same thing as a 'binary' semaphore which is the same thing as a mutex.
The main things I've seen semaphores with a count greater than one used for is producer/consumer situations in which you have a queue of a certain fixed size.
You have two semaphores then. The first semaphore is initially set to be the number of items in the queue and the second semaphore is set to 0. The producer does a P operation on the first semaphore, adds to the queue. and does a V operation on the second. The consumer does a P operation on the second semaphore, removes from the queue, and then does a V operation on the first.
In this way the producer is blocked whenever it fills the queue, and the consumer is blocked whenever the queue is empty.
A mutex is a special case of a semaphore. A semaphore allows several threads to go into the critical section. When creating a semaphore you define how may threads are allowed in the critical section. Of course your code must be able to handle several accesses to this critical section.
I find the answer of #Peer Stritzinger the correct one.
I wanted to add to his answer the following quote from the book Programming with POSIX Threads by David R Butenhof. On page 52 of chapter 3 the author writes (emphasis mine):
You cannot lock a mutex when the calling thread already has that mutex locked. The result of attempting to do so may be an error return (EDEADLK), or it may be a self-deadlock, where the unfortunate thread waits forever. You cannot unlock a mutex that is unlocked, or that is locked by another thread. Locked mutexes are owned by the thread that locks them. If you need an "unowned" lock, use a semaphore. Section 6.6.6 discusses semaphores)
With this in mind, the following piece of code illustrates the danger of using a semaphore of size 1 as a replacement for a mutex.
sem = Semaphore(1)
counter = 0 // shared variable
----
Thread 1
for (i in 1..100):
sem.lock()
++counter
sem.unlock()
----
Thread 2
for (i in 1..100):
sem.lock()
++counter
sem.unlock()
----
Thread 3
sem.unlock()
thread.sleep(1.sec)
sem.lock()
If only for threads 1 and 2, the final value of counter should be 200. However, if by mistake that semaphore reference was leaked to another thread and called unlock, than you wouldn't get mutual exclusion.
With a mutex, this behaviour would be impossible by definition.
Binary semaphore and Mutex are different. From OS perspective, a binary semaphore and counting semaphore are implemented in the same way and a binary semaphore can have a value 0 or 1.
Mutex -> Can only be used for one and only purpose of mutual exclusion for a critical section of code.
Semaphore -> Can be used to solve variety of problems. A binary semaphore can be used for signalling and also solve mutual exclusion problem. When initialized to 0, it solves signalling problem and when initialized to 1, it solves mutual exclusion problem.
When the number of resources are more and needs to be synchronized, we can use counting semaphore.
In my blog, I have discussed these topics in detail.
https://designpatterns-oo-cplusplus.blogspot.com/2015/07/synchronization-primitives-mutex-and.html