let's say we have a c++ class like:
class MyClass
{
void processArray( <an array of 255 integers> )
{
int i ;
for (i=0;i<255;i++)
{
// do something with values in the array
}
}
}
and one instance of the class like:
MyClass myInstance ;
and 2 threads which call the processArray method of that instance (depending on how system executes threads, probably in a completely irregular order). There is no mutex lock used in that scope so both threads can enter.
My question is what happens to the i ? Does each thread scope has it's own "i" or would each entering thread modify i in the for loop, causing i to be changing weirdly all the time.
i is allocated on the stack. Since each thread has its own separate stack, each thread gets its own copy of i.
Be careful. In the example provided the method processArray seems to be reentrant (it's not clear what happens in // do something with values in the array). If so, no race occurs while two or more threads invoke it simultaneously and therefore it's safe to call it without any locking mechanism.
To enforce this, you could mark both the instance and the method with the volatile qualifier, to let users know that no lock is required.
It has been published an interesting article of Andrei Alexandrescu about volatile qualifier and how it can be used to write correct multithreaded classes. The article is published here:
http://www.ddj.com/cpp/184403766
Since i is a local variable it is stored on the thread's own private stack. Hence, you do not need to protect i with a critical section.
As Adam said, i is a variable stored on the stack and the arguments are passed in so this is safe. When you have to be careful and apply mutexes or other synchronization mechanisms is if you were accessing shared member variables in the same instance of the class or global variables in the program (even scoped statics).
Related
I have a C++ class with a static function:
class Foo
{
public:
static void bar(int &a)
{
a++;
}
}
EDIT:
The variable passed as argument is only used within the calling scope. So it is not accessed by another thread.
Do i have to use a mutex when i call this function from a seperate thread?
Thanks.
Calling this function requires only thread-local resources, the stack of thread. Therefore the answer is no. If the int variable is accessible by more than the calling thread, you will need a mutex for the variable
Whether a function is static has no bearing on whether calls to it need to be synchronised.
The determining factor is whether the function is re-entrant, and what you do with the data. In this case, the function is re-entrant (as it has no non-local state of its own, or in fact any state at all), and the data is owned/managed by the calling scope, so you will have to decide within the calling scope whether that integer requires protection.
But this is true whether bar is a static member, a non-static member, a free function, a macro, a cat, a black hole, or Jon Skeet's tumble dryer.
I'd like to mention that mutex is not the only thread synchronization primitive available, and in some scenarios far from most approriate one.
Provided synchronization is needed (see two other answers on why it might be needed depending on usage) don't jump into mutex world. For something as straitghtforward as a counter (this is what I see in the code) atomic variables (or, in absence of those, atomic operations on non-atomic types) often provide a better performance and more straightforward code.
In this particular case, incrementing a variable can be easily done in thread-safe way with following C++11 code:
static void bar(std::atomic<int>& a)
{
a.fetch_add(1, std::memory_order_relaxed);
}
memory_order_relaxed used here is really far-fetched, and not neccessarily applicable (however, often good for counters). Used here mostly for the example.
I have in a Server object multiple thread who are doing the same task. Those threads are init with a Server::* routine.
In this routine there is a infinite loop with some treatments.
I was wondering if it was thread safe to use the same method for multiple threads ? No wonder for the fields of the class, If I want to read or write it I will use a mutex. But what about the routine itself ?
Since a function is an address, those thread will be running in the same memory zone ?
Do I need to create a method with same code for every thread ?
Ps: I use std::mutex(&Server::Task, this)
There is no problem with two threads running the same function at the same time (whether it's a member function or not).
In terms of instructions, it's similar to if you had two threads reading the same field at the same time - that's fine, they both get the same value. It's when you have one writing and one reading, or two writing, that you can start to have race conditions.
In C++ every thread is allocated its own call stack. This means that all local variables which exist only in the scope of a given thread's call stack belong to that thread alone. However, in the case of shared data or resources, such as a global data structure or a database, it is possible for different threads to access these at the same time. One solution to this synchronization problem is to use std::mutex, which you are already doing.
While the function itself might be the same address in memory in terms of its place in the table you aren't writing to it from multiple locations, the function itself is immutable and local variables scoped inside that function will be stacked per thread.
If your writes are protected and the fetches don't pull stale data you're as safe as you could possibly need on most architectures and implementations out there.
Behind the scenes, int Server::Task(std::string arg) is very similar to int Server__Task(Server* this, std::string arg). Just like multiple threads can execute the same function, multiple threads can also execute the same member function - even with the same arguments.
A mutex ensures that no conflicting changes are made, and that each thread sees every prior change. But since code does not chance, you don't need a mutex for it, just like you don't need a mutex for string literals.
I did find plenty of questions regarding singletons and thread safety, but none that quite answered this question for me... but I apologize if it is a repeat.
If I have a singleton object which will be used by multiple threads, I understand that any mutation to member variables should be carefully considered, but what about variables that are local to a method?
Consider this psuedo-code:
class Singleton [assume this has all the trappings of a proper singleton]
{
int doSomething() {
SomeObject obj;
obj.doStuff();
return obj.result();
}
}
In this case, is the local 'obj' thread safe? Does each thread get its own copy of it, even though there is only one object of the Singleton class?
Yes, obj is unique per thread.
There could be threading issues however if it accessed & modified common data - for example doStuff or result accesses a static member of SomeObject or some global.
As Luchian said you are fine so far, however, if you have a static or any & or * variable, then try to use a mutex lock or a spin lock. Mutex locks and spin locks exist in all unix based systems, and I think you can use them in windows as well, but you need first to add them somehow.
Here is a link for pthread mutexes: http://www.thegeekstuff.com/2012/05/c-mutex-examples/
And here another one for windows: http://msdn.microsoft.com/en-us/library/windows/desktop/ms686927(v=vs.85).aspx
Let's assume I have a simple class A with one field in C++. This field is initialized in the constructor. Class A also has a method called doit() for modifing the value of this field. doit() will be called from multiple threads. If I have a mutex only in the doit() method, is this sufficient? Do I have a guarantee that I will never read an uninitialized field (because there is no lock in the constructor)?
Edit: I probably was not clear enough. Is there no issue involving processor cache or something similar? I mean, if there is no mutex for initializing memory region (i.e. my field) - is there no risk that the other thread will read some garbage value?
Your object can only be initialised once, and you won't be able use it before it's initialised, so you don't need a mutex there. You will however need a mutex or other suitable lock in your DoIt function, as you said this will be accessed across multiple threads.
Update for edited question: No, you don't need to worry about processor cache. You must construct your object first, before you can have a handle to it. Only once you have this handle can you pass it to other threads to be used. What I'm trying to say is, the spawned threads must start after the construction of the original object, it is impossible for it to happen the other way around!
It is not possible to call doit() on an object that is not created yet, so you do not need mutex in the constructor.
If doit() is the only method that accesses the field, then you should be fine.
If other methods of your class also access that field, even from a single thread, then you must use a mutex also in these methods.
You first need to construct the object before those pesky threads get their hands on it. The OS will allocate memory for the constructor that is only called by one thread. Ths OS looks after that allocation and therefore nothing needs to be done on your part. Hell you can even create two objects of the same class in two different threads.
You can be very conservative and use a mutex at the start of any method that used that field to lock it, and release it and the end.
Or if you understand the interactions of the various methods with the various algorithms , you can use a mutex for critical sections of code that use that field - i.e. That part of the code needs to be sure that the field is not altered by another thread during processing, but you method can release the lock after the critical section, do something else then perhaps have another critical section.
I have the following sample C++ code:
class Factory
{
public:
static Factory& createInstance()
{
static Factory fac;
return fac;
}
private:
Factory()
{
//Does something non-trivial
}
};
Let's assume that createInstance is called by two threads at the same time. So will the resulting object be created properly? What happens if the second thread enters the createInstance call when the first thread is in the constructor of Factory?
C++11 and above: local static creation is thread-safe.
The standard guarantees that:
The creation is synchronized.
Should the creation throws an exception, the next time the flow of execution passes the variable definition point, creation will be attempted again.
It is generally implemented with double-checking:
first a thread-local flag is checked, and if set, then the variable is accessed.
if not yet set, then a more expensive synchronized path is taken, and if the variable is created afterward, the thread-local flag is set.
C++03 and C++98: the standard knows no thread.
There are no threads as far as the Standard is concerned, and therefore there is no provision in the Standard regarding synchronization across threads.
However some compilers implement more than the standard mandates, either in the form of extensions or by giving stronger guarantees, so check out for the compilers you're interested in. If they are good quality ones, chances are that they will guarantee it.
Finally, it might not be necessary for it to be thread-safe. If you call this method before creating any thread, then you ensures that it will be correctly initialized before the real multi-threading comes into play, and you'll neatly side-step the issue.
Looking at this page, I'd say that this is not thread-safe, because the constructor could get called multiple times before the variable is finally assigned. An InterlockedCompareExchange() might be needed, where you create a local copy of the variable, then atomically assign the pointer to a static field via the interlocked function, if the static variable is null.
Of course it's thread safe! Unless you are a complete lunatic and spawn threads from constructors of static objects, you won't have any threads until after main() is called, and the createInstance method is just returning a reference to an already constructed object, there's no way this can fail. ISO C++ guarantees that the object will be constructed before the first use after main() is called: there's no assurance that will be before main is called, but is has to be before the first use, and so all systems will perform the initialisation before main() is called. Of course ISO C++ doesn't define behaviour in the presence of threads or dynamic loading, but all compilers for host level machines provide this support and will try to preserve the semantics specified for singly threaded statically linked code where possible.
The instantiation (first call) itself is threadsafe.
However, subsequent access will not be, in general. For instance, suppose after instantiation, one thread calls a mutable Factory method and another calls some accessor method in Factory, then you will be in trouble.
For example, if your factory keeps a count of the number of instances created, you will be in trouble without some kind of mutex around that variable.
However, if Factory is truly a class with no state (no member variables), then you will be okay.