How to individually lock unordered_map elements in C++ - c++

I have an unordered_map that I want to be accessible by multiple threads but locking the whole thing with a mutex would be too slow.
To get around this I put a mutex in each element of the unordered_map:
class exampleClass{
std::mutex m;
int data;
};
std::unordered_map<int,exampleClass> exampleMap;
The issue is I'm unable to safely erase elements, because in order to destroy a mutex it must be unlocked but if it's unlocked then another thread could lock it and be writing to or reading the element during destruction.

unordered_map is not suitable for fine-grained parallelism. It is not legal
to add or remove elements without ensuring mutual exclusion during the process.
I would suggest using something like tbb::concurrent_hash_map instead, which will result in less lock contention than locking the map as a whole. (There are other concurrent hash table implementations out there; the advantage of TBB is that it's well-supported and stable.)

#Sneftel's answer is good enough.
But if you insist on using std::unordered_map, I suggest you two use one mutex to protect the insertion/deletion of the map, and another mutex for each element for modifying the element.
class exampleClass{
std::mutex m;
int data;
};
std::unordered_map<int,exampleClass> exampleMap;
std::mutex mapLock;
void add(int key, int value) {
std::unique_lock<std::mutex> _(mapLock);
exampleMap.insert({key, value});
}
void delete(int key) {
std::unique_lock<std::mutex> _(mapLock);
auto it = exampleMap.find(key);
if (it != exampleMap.end()) {
std::unique_lock<std::mutex> _1(it->m);
exampleMap.erase(it);
}
}
These should perform better for a big lock on the whole map if delete is not a frequent operation.
But be careful of these kinds of code, because it is hard to reason and to get right.
I strongly recommend #Sneftel's answer.

You have the following options:
Lock the entire mutex
Use a container of shared_ptr so the actual class can be modified (with or without a mutex) unrelated to the container.

Related

accessing to map in multithreaded environment

In my application, multiple threads need to access to a map object for inserting new items or reading the existing items. (There is no 'erase' operation).
The threads uses this code for accessing map elements:
struct PayLoad& ref = myMap[index];
I just want to know do I still need to wrap this block of this code inside of mutex ? Or is it safe to not using mutex for this purpose ?
Thanks.
Since there is at least one write operation, i.e. an insert, then you need to have thread synchronization when accessing the map. Otherwise you have a race condition.
Also, returning a reference to the value in a map is not thread-safe:
struct PayLoad& ref = myMap[index];
since multiple threads could access the value, and at least one of them could involve a write. That would also lead to a race condition. It is better to return the value by value like this:
Payload GetPayload(int index)
{
std::lock_guard<std::mutex> lock(mutex);
return myMap[index];
}
where mutex is an accessible std::mutex object.
Your insert/write operation also needs to lock the same mutex:
void SetPayload(int index, Payload payload)
{
std::lock_guard<std::mutex> lock(mutex);
myMap[index] = std::move(payload);
}

Upgrading read lock to write lock without releasing the first in C++11?

I know it's possible using boost::UpgradeLockable in C++14.
Is there anything similar for C++11?
An upgradeable lock can be written on top of simpler locking primitives.
struct upgradeable_timed_mutex {
void lock() {
upgradable_lock();
upgrade_lock();
}
void unlock() {
upgrade_unlock();
upgradable_unlock();
}
void shared_lock() { shared.shared_lock(); }
void shared_unlock() { shared.shared_unlock(); }
void upgradable_lock() { unshared.lock(); }
void ungradable_unlock() { unshared.unlock(); }
void upgrade_lock() { shared.lock(); }
void upgrade_unlock() { shared.unlock(); }
private:
friend struct upgradable_lock;
std::shared_timed_mutex shared;
std::timed_mutex unshared;
};
and similar for the timed and try variants. Note that timed variants which access two mutexes in a row have to do some extra work to avoid spending up to 2x the requested time, and try_lock has to be careful about the first lock's state in case the 2nd fails.
Then you have to write upgradable_lock, with the ability to spawn a std::unique_lock upon request.
Naturally this is hand-written thread safety code, so it is unlikely to be correct.
In C++1z you can write an untimed version as well (with std::shared_mutex and std::mutex).
Less concretely, there can be exactly one upgradeable or write lock at a time. This is what the unshared mutex represents.
So long as you hold unshared, nobody else is writing to the guarded data, so you can read from it without holding the shared mutex at all.
When you want to upgrade, you can grab a unique lock on the shared mutex. This cannot deadlock so long as no readers try to upgrade to upgradable. This excludes readers from reading, you can write, and then you release it and return back to a read only state (where you only hold the unshared mutex).

Mutex when writing to queue held in map for thread safety

I have a map<int, queue<int>> with one thread writing into it i.e. pushing messages into the queues. They key refers to a client_id, and the queue holds messages for the client. I am looking to make this read-write thread safe.
Currently, the thread that writes into it does something like this
map<int, queue<int>> msg_map;
if (msg_map.find(client_id) != msg_map.end())
{
queue<int> dummy_queue;
dummy_queue.push(msg); //msg is an int
msg_map.insert(make_pair(client_id, dummy_queue);
}
else
{
msg_map[client_id].push(msg);
}
There are many clients reading - and removing - from this map.
if (msg_map.find(client_id) != msg_map.end())
{
if (!msg_map.find(client_id)->second.empty())
{
int msg_rxed = msg_map[client_id].front();
//processing message
msg_map[client_id].pop();
}
}
I am reading this on mutexes (haven't used them before) and I was wondering when and where I ought to lock the mutex. My confusion lies in the fact that they are accessing individual queues (held within the same map). Do I lock the queues, or the map?
Is there a standard/accepted way to do this - and is using a mutex the best way to do this? There are '0s of client threads, and just that 1 single writing thread.
Simplifying and optimizing your code
For now we'll not concern ourselves with mutexes, we'll handle that later when the code is cleaned up a bit (it will be easier then).
First, from the code you showed there seems to be no reason to use an ordered std::map (logarithmic complexity), you could use the much more efficient std::unordered_map (average constant-time complexity). The choice is entirely up to you, if you don't need the container to be ordered you just have to change its declaration:
std::map<int, std::queue<int>> msg_map;
// or
std::unordered_map<int, std::queue<int>> msg_map; // C++11 only though
Now, maps are quite efficient by design but if you insist on doing lookups for each and every operation then you lose all the advantage of maps.
Concerning the writer thread, all your block of code (for the writer) can be efficiently replaced by just this line:
msg_map[client_id].push(msg);
Note that operator[] for both std::map and std::unordered_map is defined as:
Inserts a new element to the container using key as the key and a default constructed mapped value and returns a reference to the newly constructed mapped value. If an element with key key already exists, no insertion is performed and a reference to its mapped value is returned.
Concerning your reader threads, you can't directly use operator[] because it would create a new entry if none currently exists for a specific client_id so instead, you need to cache the iterator returned by find in order to reuse it and thus avoid useless lookups:
auto iter = msg_map.find(client_id);
// iter will be either std::map<int, std::queue<int>>::iterator
// or std::unordered_map<int, std::queue<int>>::iterator
if (iter != msg_map.end()) {
std::queue<int>& q = iter->second;
if (!q.empty()) {
int msg = q.front();
q.pop();
// process msg
}
}
The reason why I pop the message immediately, before processing it, is because it will improve concurrency when we add mutexes (we can unlock the mutex sooner, which is always good).
Making the code thread-safe
#hmjd's idea about multiple locks (one for the map, and one per queue) is interesting, but based on the code you showed us I disagree: any benefit you'll get from the additional concurrency will quite probably be negated by the additional time it takes to lock the queue mutexes (indeed, locking mutexes is a very expensive operation), not to mention the additional code complexity you'll have to handle. I'll bet my money on a single mutex (protecting the map and all the queues at once) being more efficient.
Incidentally, a single mutex solves the iterator invalidation problem if you want to use the more efficient std::unordered_map (std::map doesn't suffer from that problem though).
Assuming C++11, just declare a std::mutex along with your map:
std::mutex msg_map_mutex;
std::map<int, std::queue<int>> msg_map; // or std::unordered_map
Protecting the writer thread is quite straightforward, just lock the mutex before accessing the map:
std::lock_guard<std::mutex> lock(msg_map_mutex);
// the lock is held while the lock_guard object stays in scope
msg_map[client_id].push(msg);
Protecting the reader threads is barely any harder, the only trick is that you'll probably want to unlock the mutex ASAP in order to improve concurrency so you'll have to use std::unique_lock (which can be unlocked early) instead of std::lock_guard (which can only unlock when it goes out of scope):
std::unique_lock<std::mutex> lock(msg_map_mutex);
auto iter = msg_map.find(client_id);
if (iter != msg_map.end()) {
std::queue<int>& q = iter->second;
if (!q.empty()) {
int msg = q.front();
q.pop();
// assuming you don't need to access the map from now on, let's unlock
lock.unlock();
// process msg, other threads can access the map concurrently
}
}
If you can't use C++11, you'll have to replace std::mutex et al. with whatever your platform provides (pthreads, Win32, ...) or with the boost equivalent (which has the advantage of being as portable and as easy to use as the new C++11 classes, unlike the platform-specific primitives).
Read and write access to both the map and the queue need synchronized as both structures are being modified, including the map:
map<int, queue<int>> msg_map;
if (msg_map.find(client_id) != msg_map.end())
{
queue<int> dummy_queue;
dummy_queue.push(msg); //msg is an int
msg_map.insert(make_pair(client_id, dummy_queue);
}
else
{
msg_map[client_id].push(msg); // Modified here.
}
Two options are a mutex that locks both the map and queue or have a mutex for the map and a mutex per queue. The second approach is preferable as it reduces the length of time a single lock is held and means multiple threads can be updating several queues concurrently.

Is there an std or boost container that avoids race condition between its insert and find methods?

Each thread may insert an object to the container (at most once) using 'insert' function.
Hereafter, the thread may try to access this object using 'get' function.
Therefore, there is no race between 'insert' and 'get' when used by the same thread.
However, a different thread may try to insert its own object while another thread has called 'get'
I need a container where this situation does not need any synchronization method.
The number of threads may vary dramatically between executions.
class Object;
class Container<Object>;
Container<Object> g_container;
void insert(int threadId)
{
ScopedLock<Mutex> lock(insertMutex);
Object obj;
g_container[threadId] = obj;
}
Object get(int threadId)
{
return g_container[threadId];
}
You can use a vector of container pointers. Each thread manages their own container instance, and registers it with the array of container pointers.
template <typename T>
struct Container {
Mutex insertMutex;
ContainerType<T> container;
int index;
void insert (T &obj) {
ScopedLock<Mutex> lock(insertMutex);
container.insert(obj);
}
T get () {
return *container.begin();
}
void register_container () {
if (index != 0) return;
if (counter == MAX_THREADS) throw 0;
index = ++counter;
ScopedLock<Mutex> lock(registerMutex);
containers.push_back(this);
}
static std::vector<Container *> containers;
static Atomic<int> counter;
static Mutex registerMutex;
};
template <typename T>
std::vector<Container<T> *> Container<T>::containers;
template <typename T>
Atomic<int> Container<T>::counter;
template <typename T>
Mutex Container<T>::registerMutex;
Now, you can iterate over the Container<T>::containers to access all containers.
The thing you're looking for is usually called a "lock-free data structure".
I understand why you think you need a lock-free container, but I can almost promise you that you don't. They are usually more trouble than they're worth. You should just have one mutex which governs all access to an ordinary container. Lock the mutex to insert in the container. Lock the mutex to remove from the container. And lock the mutex before reading (or iterating over) the container.
Here is some helpful discussion from Herb Sutter, chair of the C++ standards committee:
http://www.drdobbs.com/cpp/lock-free-code-a-false-sense-of-security/210600279
+1 to James Brock.
But it sounds more like you want one container for each thread, yet mutexed, and locking/unlocking the mutexes is cheap because 99.9% of the time the operation succeeds the first time. But the "master" thread does have the ability to occasionally pause the other threads while examining them.
If each thread has only one object, the containers go away. Just lock the objects. Don't worry about the cost of locking unless it's a proven performance problem. In that case, the problem isn't to remove the locks, but merely to reduce their usage frequency. So keep the objects locked and only do an unlock every hundredth time or so. Combine with a condition variable if necessary.

Is it possible to use mutex to lock only one element of a data structure ?

Is it possible to use mutex to lock only one element of a data structure ?
e.g.
boost::mutex m_mutex;
map<string, int> myMap;
// initialize myMap so that it has 10 elements
// then in thread 1
{
boost::unique_lock<boost::mutex> lock(m_mutex);
myMap[1] = 5 ; // write map[1]
}
// in thread 2
{
boost::unique_lock<boost::mutex> lock(m_mutex);
myMap[2] = 4 ; // write map[1]
}
My question:
When thread 1 is writing map[1], thread 2 can writing map[2] at the same time ?
The thread lock the whole map data structure or only an element, e.g. map[1] or map[2].
thanks
If you can guarantee that nobody is modifying the container itself (via insert and erase etc.), then as long as each thread accesses a different element of the container, you should be fine.
If you need per-element locking, you could modify the element type to something that offers synchronized access. (Worst case a pair of a mutex and the original value.)
You need a different mutex for every element of the map. You can do this with a map of mutex or adding a mutex to the mapped type (in your case it is int, so you can't do it without creating a new class like SharedInt)
Mutexes lock executable regions not objects. I always think about locking any code regions that read/modify thread objects. If an object is locked within a region but that object is accessible within another un-synchronized code region, you are not safe (ofcourse). In your case, I'd lock access to the entire object as insertions and reading from containers can easily experience context switching and thus increase the likelihood of data corruption.
Mutex is all about discipline. One thread can call write and other thread can call write1. C++ runtime will assume it is intentional. But most of the cases it is not the programmer intended. Summary is as long as all threads/methods follow the discipline (understand the the critical section and respect it) there will be consistency.
int i=0;
Write()
{
//Lock
i++;
//Unlock
}
Write1()
{
i++;
}