I'm referring to this example.
The authors use memory_order_release to decrement the counter. And they even state in the discussion section that using memory_order_acq_rel instead would be excessive. But wouldn't the following scenario in theory lead to that x is never deleted?
we have two threads on different CPUs
each of them owns an instance of a shared pointer, both pointers share ownership over the same control block, no other pointers referring that block exist
each thread has the counter in its cache and the counter is 2 for both of them
the first thread destroys its pointer, the counter in this thread now is 1
the second thread destroys its pointer, however cache invalidation signal from the first thread may still be queued, so it decrements the value from its own cache and gets 1
both threads didn't delete x, but there are no shared pointers sharing our control block => memory leak
The code sample from the link:
#include <boost/intrusive_ptr.hpp>
#include <boost/atomic.hpp>
class X {
public:
typedef boost::intrusive_ptr<X> pointer;
X() : refcount_(0) {}
private:
mutable boost::atomic<int> refcount_;
friend void intrusive_ptr_add_ref(const X * x)
{
x->refcount_.fetch_add(1, boost::memory_order_relaxed);
}
friend void intrusive_ptr_release(const X * x)
{
if (x->refcount_.fetch_sub(1, boost::memory_order_release) == 1) {
boost::atomic_thread_fence(boost::memory_order_acquire);
delete x;
}
}
};
The quote from discussion section:
It would be possible to use memory_order_acq_rel for the fetch_sub
operation, but this results in unneeded "acquire" operations when the
reference counter does not yet reach zero and may impose a performance
penalty.
All modifications of a single atomic variable happen in a global modification order. It is not possible for two threads to disagree about this order.
The fetch_sub operation is an atomic read-modify-write operation and is required to always read the value of the atomic variable immediately before the modification from the same operation in the modification order.
So it is not possible for the second thread to read 2 when the first thread's fetch_sub was first in the modification order. The implementation must assure that such a cache incoherence cannot happen, if necessary with the help of locks if the hardware doesn't support this atomic access natively. (That is what the is_lock_free and is_always_lock_free members of the atomic are there to check for.)
This is all independent of the memory orders of the operations. These matter only for access to other memory locations than the atomic variable itself.
Related
I have a global reference-counted object obj that I want to protect from data races by using atomic operations:
T* obj; // initially nullptr
std::atomic<int> count; // initially zero
My understanding is that I need to use std::memory_order_release after I write to obj, so that the other threads will be aware of it being created:
void increment()
{
if (count.load(std::memory_order_relaxed) == 0)
obj = std::make_unique<T>();
count.fetch_add(1, std::memory_order_release);
}
Likewise, I need to use std::memory_order_acquire when reading the counter, to ensure the thread has visibility of obj being changed:
void decrement()
{
count.fetch_sub(1, std::memory_order_relaxed);
if (count.load(std::memory_order_acquire) == 0)
obj.reset();
}
I am not convinced that the code above is correct, but I'm not entirely sure why. I feel like after obj.reset() is called, there should be a std::memory_order_release operation to inform other threads about it. Is that correct?
Are there other things that can go wrong, or is my understanding of atomic operations in this case completely wrong?
It is wrong regardless of memory ordering.
As #MaartenBamelis pointed out for concurrent calling of increment the object is constructed twice. And the same is true for concurrent decrement: object is reset twice (which may result in double destructor call).
Note that there's disagreement between T* obj; declaration and using it as unique_ptr but neither raw pointer not unique pointer are safe for concurrent modification. In practice, reset or delete will check pointer for null, then delete and set it to null, and these steps are not atomic.
fetch_add and fetch_sub are fetch and op instead of just op for a reason: if you don't use the value observed during operation, it is likely to be a race.
This code is inherently racey. If two threads call increment at the same time when count is initially 0, both will see count as 0, and both will create obj (and race to see which copy is kept; given unique_ptr has no special threading protections, terrible things can happen if two of them set it at once).
If two threads decrement at the same time (holding the last two references), and finish the fetch_sub before either calls load, both will reset obj (also bad).
And if a decrement finishes the fetch_sub (to 0), then another thread increments before the decrement load occurs, the increment will see the count as 0 and reinitialize. Whether the object is cleared after being replaced, or replaced after being cleared, or some horrible mixture of the two, will depend on whether increment's fetch_add runs before or after decrement's load.
In short: If you find yourself using two separate atomic operations on the same variable, and testing the result of one of them (without looping, as in a compare and swap loop), you're wrong.
More correct code would look like:
void increment() // Still not safe
{
// acquire is good for the != 0 case, for a later read of obj
// or would be if the other writer did a release *after* constructing an obj
if (count.fetch_add(1, std::memory_order_acquire) == 0)
obj = std::make_unique<T>();
}
void decrement()
{
if (count.fetch_sub(1, std::memory_order_acquire) == 1)
obj.reset();
}
but even then it's not reliable; there's no guarantee that, when count is 0, two threads couldn't call increment, both of them fetch_add at once, and while exactly one of them is guaranteed to see the count as 0, said 0-seeing thread might end up delayed while the one that saw it as 1 assumes the object exists and uses it before it's initialized.
I'm not going to swear there's no mutex-free solution here, but dealing with the issues involved with atomics is almost certainly not worth the headache.
It might be possible to confine the mutex to inside the if() branches, but taking a mutex is also an atomic RMW operation (and not much more than that for a good lightweight implementation) so this doesn't necessarily help a huge amount. If you need really good read-side scaling, you'd want to look into something like RCU instead of a ref-count, to allow readers to truly be read-only, not contending with other readers.
I don't really see a simple way of implementing a reference-counted resource with atomics. Maybe there's some clever way that I haven't thought of yet, but in my experience, clever does not equal readable.
My advice would be to implement it first using a mutex. Then you simply lock the mutex, check the reference count, do whatever needs to be done, and unlock again. It's guaranteed correct:
std::mutex mutex;
int count;
std::unique_ptr<T> obj;
void increment()
{
auto lock = std::scoped_lock{mutex};
if (++count == 1) // Am I the first reference?
obj = std::make_unique<T>();
}
void decrement()
{
auto lock = std::scoped_lock{mutex};
if (--count == 0) // Was I the last reference?
obj.reset();
}
Although at this point, I would just use a std::shared_ptr instead of managing the reference count myself:
std::mutex mutex;
std::weak_ptr<T> obj;
std::shared_ptr<T> acquire()
{
auto lock = std::scoped_lock{mutex};
auto sp = obj.lock();
if (!sp)
obj = sp = std::make_shared<T>();
return sp;
}
I believe this also makes it safe when exceptions may be thrown when constructing the object.
Mutexes are surprisingly performant, so I expect that locking code is plenty quick unless you have a highly specialized use case where you need code to be lock-free.
I am working on a small program that utilizes shared pointers. I have a simple class "Thing", that just is an class with an integer attribute:
class Thing{
public:
Thing(int m){x=m;}
int operator()(){
return x;
}
void set_val(int v){
x=v;
}
int x;
~Thing(){
std::cout<<"Deleted thing with value "<<x<<std::endl;
}
};
I have a simple function "fun", that takes in a shared_ptr instance, and an integer value, index, which just keeps track of which thread is outputting a given value. The function prints out the index value, passed to the function, and the reference count of the shared pointer that was passed in as an argument
std::mutex mtx1;
void fun(std::shared_ptr<Thing> t1,int index){
std::lock_guard <std::mutex> loc(mtx1);
int m=t1.use_count();
std::cout<<index<<" : "<<m<<std::endl;
}
In main,I create one instance of a shared pointer which is a wrapper around a Thing object like so:
std::shared_ptr<Thing> ptr5(nullptr);
ptr5=std::make_shared<Thing>(110);
(declared this way for exception safety).
I then create 3 threads, each of which creates a thread executing the fun() function that takes in as arguments the ptr5 shared pointer and increasing index values:
std::thread t1(fun,ptr5,1),t2(fun,ptr5,2),t3(fun,ptr5,3);
t1.join();
t2.join();
t3.join();
My thought process here is that since each of the shared pointer control block member functions was thread safe, the call to use_count() within the fun() function is not an atomic instruction and therefore requires no locking. However, both running without and without a lock_guard,this still leads to a race condition. I expect to see an output of:
1:2
2:3
3:4
since each thread spawns a new reference to the original shared pointer, the use_count() will be incremented each time by 1 reference. However, my output is still random due to some race condition.
In a multithreaded environment, use_count() is approximate. From cppreference :
In multithreaded environment, the value returned by use_count is approximate (typical implementations use a memory_order_relaxed load)
The control block for shared_ptr is otherwise thread-safe :
All member functions (including copy constructor and copy assignment) can be called by multiple threads on different instances of shared_ptr without additional synchronization even if these instances are copies and share ownership of the same object.
Seeing an out-of-date use_count() is not an indication that the control-block was corrupted by a race condition.
Note that this does not extend to modifying the pointed object. It does not provide synchronization for the pointed object. Only the state of the shared_ptr and the control block are protected.
This article by Jeff Preshing states that the double-checked locking pattern (DCLP) is fixed in C++11. The classical example used for this pattern is the singleton pattern but I happen to have a different use case and I am still lacking experience in handling "atomic<> weapons" - maybe someone over here can help me out.
Is the following piece of code a correct DCLP implementation as described by Jeff under "Using C++11 Sequentially Consistent Atomics"?
class Foo {
std::shared_ptr<B> data;
std::mutex mutex;
void detach()
{
if (data.use_count() > 1)
{
std::lock_guard<std::mutex> lock{mutex};
if (data.use_count() > 1)
{
data = std::make_shared<B>(*data);
}
}
}
public:
// public interface
};
No, this is not a correct implementation of DCLP.
The thing is that your outer check data.use_count() > 1 accesses the object (of type B with reference count), which can be deleted (unreferenced) in mutex-protected part. Any sort of memory fences cannot help there.
Why data.use_count() accesses the object:
Assume these operations have been executed:
shared_ptr<B> data1 = make_shared<B>(...);
shared_ptr<B> data = data1;
Then you have following layout (weak_ptr support is not shown here):
data1 [allocated with B::new()] data
--------------------------
[pointer type] ref; --> |atomic<int> m_use_count;| <-- [pointer type] ref
|B obj; |
--------------------------
Each shared_ptr object is just a pointer, which points to allocated memory region. This memory region embeds object of type B plus atomic counter, reflecting number of shared_ptr's, pointed to given object. When this counter becomes zero, memory region is freed(and B object is destroyed). Exactly this counter is returned by shared_ptr::use_count().
UPDATE: Execution, which can lead to accessing memory which is already freed (initially, two shared_ptr's point to the same object, .use_count() is 2):
/* Thread 1 */ /* Thread 2 */ /* Thread 3 */
Enter detach() Enter detach()
Found `data.use_count()` > 1
Enter critical section
Found `data.use_count()` > 1
Dereference `data`,
found old object.
Unreference old `data`,
`use_count` becomes 1
Delete other shared_ptr,
old object is deleted
Assign new object to `data`
Access old object
(for check `use_count`)
!! But object is freed !!
Outer check should only take a pointer to object for decide, whether to need aquire lock.
BTW, even your implementation would be correct, it has a little sence:
If data (and detach) can be accessed from several threads at the same time, object's uniqueness gives no advantages, since it can be accessed from the several threads. If you want to change object, all accesses to data should be protected by outer mutex, in that case detach() cannot be executed concurrently.
If data (and detach) can be accessed only by single thread at the same time, detach implementation doesn't need any locking at all.
This constitutes a data race if two threads invoke detach on the same instance of Foo concurrently, because std::shared_ptr<B>::use_count() (a read-only operation) would run concurrently with the std::shared_ptr<B> move-assignment operator (a modifying operation), which is a data race and hence a cause of undefined behavior. If Foo instances are never accessed concurrently, on the other hand, there is no data race, but then the std::mutex would be useless in your example. The question is: how does data's pointee become shared in the first place? Without this crucial bit of information, it is hard to tell if the code is safe even if a Foo is never used concurrently.
According to your source, I think you still need to add thread fences before the first test and after the second test.
std::shared_ptr<B> data;
std::mutex mutex;
void detach()
{
std::atomic_thread_fence(std::memory_order_acquire);
if (data.use_count() > 1)
{
auto lock = std::lock_guard<std::mutex>{mutex};
if (data.use_count() > 1)
{
std::atomic_thread_fence(std::memory_order_release);
data = std::make_shared<B>(*data);
}
}
}
Assuming that aligned pointer loads and stores are naturally atomic on the target platform, what is the difference between this:
// Case 1: Dumb pointer, manual fence
int* ptr;
// ...
std::atomic_thread_fence(std::memory_order_release);
ptr = new int(-4);
this:
// Case 2: atomic var, automatic fence
std::atomic<int*> ptr;
// ...
ptr.store(new int(-4), std::memory_order_release);
and this:
// Case 3: atomic var, manual fence
std::atomic<int*> ptr;
// ...
std::atomic_thread_fence(std::memory_order_release);
ptr.store(new int(-4), std::memory_order_relaxed);
I was under the impression that they were all equivalent, however Relacy detects a data race in
the first case (only):
struct test_relacy_behaviour : public rl::test_suite<test_relacy_behaviour, 2>
{
rl::var<std::string*> ptr;
rl::var<int> data;
void before()
{
ptr($) = nullptr;
rl::atomic_thread_fence(rl::memory_order_seq_cst);
}
void thread(unsigned int id)
{
if (id == 0) {
std::string* p = new std::string("Hello");
data($) = 42;
rl::atomic_thread_fence(rl::memory_order_release);
ptr($) = p;
}
else {
std::string* p2 = ptr($); // <-- Test fails here after the first thread completely finishes executing (no contention)
rl::atomic_thread_fence(rl::memory_order_acquire);
RL_ASSERT(!p2 || *p2 == "Hello" && data($) == 42);
}
}
void after()
{
delete ptr($);
}
};
I contacted the author of Relacy to find out if this was expected behaviour; he says that there is indeed a data race in my test case.
However, I'm having trouble spotting it; can someone point out to me what the race is?
Most importantly, what are the differences between these three cases?
Update: It's occurred to me that Relacy may simply be complaining about the atomicity (or lack thereof, rather) of the variable being accessed across threads... after all, it doesn't know that I intend only to use this code on platforms where aligned integer/pointer access is naturally atomic.
Another update: Jeff Preshing has written an excellent blog post explaining the difference between explicit fences and the built-in ones ("fences" vs "operations"). Cases 2 and 3 are apparently not equivalent! (In certain subtle circumstances, anyway.)
Although various answers cover bits and pieces of what the potential problem is and/or provide useful information, no answer correctly describes the potential issues for all three cases.
In order to synchronize memory operations between threads, release and acquire barriers are used to specify ordering.
In the diagram, memory operations A in thread 1 cannot move down across the (one-way) release barrier (regardless whether that is a release operation on an atomic store,
or a standalone release fence followed by a relaxed atomic store). Hence memory operations A are guaranteed to happen before the atomic store.
Same goes for memory operations B in thread 2 which cannot move up across the acquire barrier; hence the atomic load happens before memory operations B.
The atomic ptr itself provides inter-thread ordering based on the guarantee that it has a single modification order. As soon as thread 2 sees a value for ptr,
it is guaranteed that the store (and thus memory operations A) happened before the load. Because the load is guaranteed to happen before memory operations B,
the rules for transitivity say that memory operations A happen before B and synchronization is complete.
With that, let's look at your 3 cases.
Case 1 is broken because ptr, a non-atomic type, is modified in different threads. That is a classical example of a data race and it causes undefined behavior.
Case 2 is correct. As an argument, the integer allocation with new is sequenced before the release operation. This is equivalent to:
// Case 2: atomic var, automatic fence
std::atomic<int*> ptr;
// ...
int *tmp = new int(-4);
ptr.store(tmp, std::memory_order_release);
Case 3 is broken, albeit in a subtle way. The problem is that even though the ptr assignment is correctly sequenced after the standalone fence,
the integer allocation (new) is also sequenced after the fence, causing a data race on the integer memory location.
the code is equivalent to:
// Case 3: atomic var, manual fence
std::atomic<int*> ptr;
// ...
std::atomic_thread_fence(std::memory_order_release);
int *tmp = new int(-4);
ptr.store(tmp, std::memory_order_relaxed);
If you map that to the diagram above, the new operator is supposed to be part of memory operations A. Being sequenced below the release fence,
ordering guarantees no longer hold and the integer allocation may actually be reordered with memory operations B in thread 2.
Therefore, a load() in thread 2 may return garbage or cause other undefined behavior.
I believe the code has a race. Case 1 and case 2 are not equivalent.
29.8 [atomics.fences]
-2- A release fence A synchronizes with an acquire fence B if there exist atomic operations X and Y, both operating on some atomic object M, such that A is sequenced before X, X modifies M, Y is sequenced before B, and Y reads the value written by X or a value written by any side effect in the hypothetical release sequence X would head if it were a release operation.
In case 1 your release fence does not synchronize with your acquire fence because ptr is not an atomic object and the store and load on ptr are not atomic operations.
Case 2 and case 3 are equivalent (actually, not quite, see LWimsey's comments and answer), because ptr is an atomic object and the store is an atomic operation. (Paragraphs 3 and 4 of [atomic.fences] describe how a fence synchronizes with an atomic operation and vice versa.)
The semantics of fences are defined only with respect to atomic objects and atomic operations. Whether your target platform and your implementation offer stronger guarantees (such as treating any pointer type as an atomic object) is implementation-defined at best.
N.B. for both of case 2 and case 3 the acquire operation on ptr could happen before the store, and so would read garbage from the uninitialized atomic<int*>. Simply using acquire and release operations (or fences) doesn't ensure that the store happens before the load, it only ensures that if the load reads the stored value then the code is correctly synchronized.
Several pertinent references:
the C++11 draft standard (PDF, see clauses 1, 29 and 30);
Hans-J. Boehm's overview of concurrency in C++;
McKenney, Boehm and Crowl on concurrency in C++;
GCC's developmental notes on concurrency in C++;
the Linux kernel's notes on concurrency;
a related question with answers here on Stackoverflow;
another related question with answers;
Cppmem, a sandbox in which to experiment with concurrency;
Cppmem's help page;
Spin, a tool for analyzing the logical consistency of concurrent systems;
an overview of memory barriers from a hardware perspective (PDF).
Some of the above may interest you and other readers.
The memory backing an atomic variable can only ever be used for the contents of the atomic. However, a plain variable, like ptr in case 1, is a different story. Once a compiler has the right to write to it, it can write anything to it, even the value of a temporary value when you run out of registers.
Remember, your example is pathologically clean. Given a slightly more complex example:
std::string* p = new std::string("Hello");
data($) = 42;
rl::atomic_thread_fence(rl::memory_order_release);
std::string* p2 = new std::string("Bye");
ptr($) = p;
it is totally legal for the compiler to choose to reuse your pointer
std::string* p = new std::string("Hello");
data($) = 42;
rl::atomic_thread_fence(rl::memory_order_release);
ptr($) = new std::string("Bye");
std::string* p2 = ptr($);
ptr($) = p;
Why would it do so? I don't know, perhaps some exotic trick to keep a cache line or something. The point is that, since ptr is not atomic in case 1, there is a race case between the write on line 'ptr($) = p' and the read on 'std::string* p2 = ptr($)', yielding undefined behavior. In this simple test case, the compiler may not choose to exercise this right, and it may be safe, but in more complicated cases the compiler has the right to abuse ptr however it pleases, and Relacy catches this.
My favorite article on the topic: http://software.intel.com/en-us/blogs/2013/01/06/benign-data-races-what-could-possibly-go-wrong
The race in the first example is between the publication of the pointer, and the stuff that it points to. The reason is, that you have the creation and initialization of the pointer after the fence (= on the same side as the publication of the pointer):
int* ptr; //noop
std::atomic_thread_fence(std::memory_order_release); //fence between noop and interesting stuff
ptr = new int(-4); //object creation, initalization, and publication
If we assume that CPU accesses to properly aligned pointers are atomic, the code can be corrected by writing this:
int* ptr; //noop
int* newPtr = new int(-4); //object creation & initalization
std::atomic_thread_fence(std::memory_order_release); //fence between initialization and publication
ptr = newPtr; //publication
Note that even though this may work fine on many machines, there is absolutely no guarantee within the C++ standard on the atomicity of the last line. So better use atomic<> variables in the first place.
Derived from this question and related to this question:
If I construct an object in one thread and then convey a reference/pointer to it to another thread, is it thread un-safe for that other thread to access the object without explicit locking/memory-barriers?
// thread 1
Obj obj;
anyLeagalTransferDevice.Send(&obj);
while(1); // never let obj go out of scope
// thread 2
anyLeagalTransferDevice.Get()->SomeFn();
Alternatively: is there any legal way to convey data between threads that doesn't enforce memory ordering with regards to everything else the thread has touched? From a hardware standpoint I don't see any reason it shouldn't be possible.
To clarify; the question is with regards to cache coherency, memory ordering and whatnot. Can Thread 2 get and use the pointer before Thread 2's view of memory includes the writes involved in constructing obj? To miss-quote Alexandrescu(?) "Could a malicious CPU designer and compiler writer collude to build a standard conforming system that make that break?"
Reasoning about thread-safety can be difficult, and I am no expert on the C++11 memory model. Fortunately, however, your example is very simple. I rewrite the example, because the constructor is irrelevant.
Simplified Example
Question: Is the following code correct? Or can the execution result in undefined behavior?
// Legal transfer of pointer to int without data race.
// The receive function blocks until send is called.
void send(int*);
int* receive();
// --- thread A ---
/* A1 */ int* pointer = receive();
/* A2 */ int answer = *pointer;
// --- thread B ---
int answer;
/* B1 */ answer = 42;
/* B2 */ send(&answer);
// wait forever
Answer: There may be a data race on the memory location of answer, and thus the execution results in undefined behavior. See below for details.
Implementation of Data Transfer
Of course, the answer depends on the possible and legal implementations of the functions send and receive. I use the following data-race-free implementation. Note that only a single atomic variable is used, and all memory operations use std::memory_order_relaxed. Basically this means, that these functions do not restrict memory re-orderings.
std::atomic<int*> transfer{nullptr};
void send(int* pointer) {
transfer.store(pointer, std::memory_order_relaxed);
}
int* receive() {
while (transfer.load(std::memory_order_relaxed) == nullptr) { }
return transfer.load(std::memory_order_relaxed);
}
Order of Memory Operations
On multicore systems, a thread can see memory changes in a different order as what other threads see. In addition, both compilers and CPUs may reorder memory operations within a single thread for efficiency - and they do this all the time. Atomic operations with std::memory_order_relaxed do not participate in any synchronization and do not impose any ordering.
In the above example, the compiler is allowed to reorder the operations of thread B, and execute B2 before B1, because the reordering has no effect on the thread itself.
// --- valid execution of operations in thread B ---
int answer;
/* B2 */ send(&answer);
/* B1 */ answer = 42;
// wait forever
Data Race
C++11 defines a data race as follows (N3290 C++11 Draft): "The execution of a program contains a data race if it contains two conflicting actions in different threads, at least one of which is not atomic, and neither happens before the other. Any such data race results in undefined behavior." And the term happens before is defined earlier in the same document.
In the above example, B1 and A2 are conflicting and non-atomic operations, and neither happens before the other. This is obvious, because I have shown in the previous section, that both can happen at the same time.
That's the only thing that matters in C++11. In contrast, the Java Memory Model also tries to define the behavior if there are data races, and it took them almost a decade to come up with a reasonable specification. C++11 didn't make the same mistake.
Further Information
I'm a bit surprised that these basics are not well known. The definitive source of information is the section Multi-threaded executions and data races in the C++11 standard. However, the specification is difficult to understand.
A good starting point are Hans Boehm's talks - e.g. available as online videos:
Threads and Shared Variables in C++11
Getting C++ Threads Right
There are also a lot of other good resources, I have mentioned elsewhere, e.g.:
std::memory_order - cppreference.com
There is no parallel access to the same data, so there is no problem:
Thread 1 starts execution of Obj::Obj().
Thread 1 finishes execution of Obj::Obj().
Thread 1 passes reference to the memory occupied by obj to thread 2.
Thread 1 never does anything else with that memory (soon after, it falls into infinite loop).
Thread 2 picks-up the reference to memory occupied by obj.
Thread 2 presumably does something with it, undisturbed by thread 1 which is still infinitely looping.
The only potential problem is if Send didn't acts as a memory barrier, but then it wouldn't really be a "legal transfer device".
As others have alluded to, the only way in which a constructor is not thread-safe is if something somehow gets a pointer or reference to it before the constructor is finished, and the only way that would occur is if the constructor itself has code that registers the this pointer to some type of container which is shared across threads.
Now in your specific example, Branko Dimitrijevic gave a good complete explanation how your case is fine. But in the general case, I'd say to not use something until the constructor is finished, though I don't think there's anything "special" that doesn't happen until the constructor is finished. By the time it enters the (last) constructor in an inheritance chain, the object is pretty much fully "good to go" with all of its member variables being initialized, etc. So no worse than any other critical section work, but another thread would need to know about it first, and the only way that happens is if you're sharing this in the constructor itself somehow. So only do that as the "last thing" if you are.
It is only safe (sort of) if you wrote both threads, and know the first thread is not accessing it while the second thread is. For example, if the thread constructing it never accesses it after passing the reference/pointer, you would be OK. Otherwise it is thread unsafe. You could change that by making all methods that access data members (read or write) lock memory.
Read this question until now... Still will post my comments:
Static Local Variable
There is a reliable way to construct objects when you are in a multi-thread environment, that is using a static local variable (static local variable-CppCoreGuidelines),
From the above reference: "This is one of the most effective solutions to problems related to initialization order. In a multi-threaded environment the initialization of the static object does not introduce a race condition (unless you carelessly access a shared object from within its constructor)."
Also note from the reference, if the destruction of X involves an operation that needs to be synchronized you can create the object on the heap and synchronize when to call the destructor.
Below is an example I wrote to show the Construct On First Use Idiom, which is basically what the reference talks about.
#include <iostream>
#include <thread>
#include <vector>
class ThreadConstruct
{
public:
ThreadConstruct(int a, float b) : _a{a}, _b{b}
{
std::cout << "ThreadConstruct construct start" << std::endl;
std::this_thread::sleep_for(std::chrono::seconds(2));
std::cout << "ThreadConstruct construct end" << std::endl;
}
void get()
{
std::cout << _a << " " << _b << std::endl;
}
private:
int _a;
float _b;
};
struct Factory
{
template<class T, typename ...ARGS>
static T& get(ARGS... args)
{
//thread safe object instantiation
static T instance(std::forward<ARGS>(args)...);
return instance;
}
};
//thread pool
class Threads
{
public:
Threads()
{
for (size_t num_threads = 0; num_threads < 5; ++num_threads) {
thread_pool.emplace_back(&Threads::run, this);
}
}
void run()
{
//thread safe constructor call
ThreadConstruct& thread_construct = Factory::get<ThreadConstruct>(5, 10.1);
thread_construct.get();
}
~Threads()
{
for(auto& x : thread_pool) {
if(x.joinable()) {
x.join();
}
}
}
private:
std::vector<std::thread> thread_pool;
};
int main()
{
Threads thread;
return 0;
}
Output:
ThreadConstruct construct start
ThreadConstruct construct end
5 10.1
5 10.1
5 10.1
5 10.1
5 10.1