Before coroutines we used callbacks to run asynchronous operations. Callbacks was normal functions and could have thread_local variables.
Let see this example:
void StartTcpConnection(void)
{
using namespace std;
thread_local int my_thread_local = 1;
cout << "my_thread_local = " << my_thread_local << endl;
auto tcp_connection = tcp_connect("127.0.0.1", 8080);
tcp_connection.async_wait(TcpConnected);
}
void TcpConnected(void)
{
using namespace std;
thread_local int my_thread_local = 2;
cout << "my_thread_local = " << my_thread_local << endl;
}
As we see from code, I have some (undocumented here) tcp_connect function that connects to TCP endpoint and returns tcp_connection object. This object can wait until TCP connection will really occur and call TcpConnected function. Because we don't know specific implementation of tcp_connect and tcp_connection, we don't know will it call TcpConnected on the same or on different thread, both implementations are possible. But we know for sure that my_thread_local is different for 2 different functions, because each function has its own scope.
If we need this variable to be the same (as soon as thread is the same), we can create 3rd function that will return reference to thread_local variable:
int& GetThreadLocalInt(void)
{
thread_local int my_variable = 1;
return my_variable;
}
So, we have full control and predictability: we know for sure that variables will be different if TcpConnected and StartTcpConnection will run on different threads, and we know that we can have them different or the same depending on our choice when these functions will run on the same thread.
Now let see coroutine version of the same operation:
void Tcp(void)
{
thread_local int my_thread_local = 1;
auto tcp_connection = co_await tcp_connect("127.0.0.1", 8080);
cout << "my_thread_local = " << my_thread_local << endl;
}
This situation is a bit questionable for me. I still need thread local storage, it is important language feature that I don't want to abandon. However, we here have 2 cases:
Thread before co_await is the same one as after co_await. What will happen with my_thread_local? Will it be the same variable before and after co_await, especially if we'll use GetThreadLocalInt function to get its reference instead of value?
Thread changes after co_await. Will C++ runtime reinitialize my_thread_local to value from new thread, or make a copy of previous thread value, or may be use reference to the same data? And similar question for GetThreadLocalInt function, it returns reference to thread_local object, but the reference storage itself is auto, will coroutine reinitialize it to new thread, or we'll get (dangerous!!!) race condition, because thread 2 will strangely get reference to thread 1 thread local data and potentially use it in parallel?
Even it is easy to debug and test what will happen on any specific compiler, the important question is whether standard says us something about that, otherwise even if we'll test it on VC++ or gcc an see that it behaves somehow on these 2 popular compilers, the code may loose portability and compile differently on some exotic compilers.
For global thread_local variables, the coroutine behavior ought to be as expected (and MSVC seems to have a bug with this). But function-local thread_local variables in coroutines, there seems to be a hole in the specification. Indeed, I'm not sure the wording makes sense even without coroutines.
[stmt.dcl]/3 says:
Dynamic initialization of a block variable with static storage duration or thread storage duration is performed the first time control passes through its declaration; such a variable is considered initialized upon the completion of its initialization.
The problem is that, while by C++'s rules there is only one thread_local variable, there are multiple objects represented by that one variable. And those objects need to be initialized. So... how does that happen?
The only sane interpretation of this is that the object for a particular thread gets initialized the first time control flow passes through the declaration on that thread. And that's the problem.
If using co_await the thread a function executes on changes, then how could you pass through the declaration on the new thread? Which means that the thread_local for that thread should be zero-initialized.
Ultimately, I would say that you should never use thread_local in a coroutine function. It's just not clear what value the thread_local ought to have. And the only logical value for the new thread_local to have is the one from the previous thread. But that's not how a thread_local is supposed to work. Overall, the idea feels inherently nonsensical (and I would say that the standard should have explicitly forbid the declaration of thread_locals in a coroutine, just as they did for using co_await in a thread_local's initializer).
Just use a namespace-scoped thread_local variable. Outside of the aforementioned MSVC bug, it ought to work and make sense.
Related
I am confused with the description of thread_local in C++11. My understanding is, each thread has unique copy of local variables in a function. The global/static variables can be accessed by all the threads (possibly synchronized access using locks). And the thread_local variables are visible to all the threads but can only modified by the thread for which they are defined? Is it correct?
Thread-local storage duration is a term used to refer to data that is seemingly global or static storage duration (from the viewpoint of the functions using it) but, in actual fact, there is one copy per thread.
It adds to the current options:
automatic (exists during a block or function);
static (exists for the program duration); and
dynamic (exists on the heap between allocation and deallocation).
Something that is thread-local is brought into existence at thread creation time and disposed of when the thread finishes.
For example, think of a random number generator where the seed must be maintained on a per-thread basis. Using a thread-local seed means that each thread gets its own random number sequence, independent of all other threads.
If your seed was a local variable within the random function, it would be initialised every time you called it, giving you the same number each time. If it was a global, threads would interfere with each other's sequences.
Another example is something like strtok where the tokenisation state is stored on a thread-specific basis. That way, a single thread can be sure that other threads won't screw up its tokenisation efforts, while still being able to maintain state over multiple calls to strtok - this basically renders strtok_r (the thread-safe version) redundant.
Yet another example would be something like errno. You don't want separate threads modifying errno after one of your calls fails, but before you've had a chance to check the result.
This site has a reasonable description of the different storage duration specifiers.
When you declare a variable thread_local then each thread has its own copy. When you refer to it by name, then the copy associated with the current thread is used. e.g.
thread_local int i=0;
void f(int newval){
i=newval;
}
void g(){
std::cout<<i;
}
void threadfunc(int id){
f(id);
++i;
g();
}
int main(){
i=9;
std::thread t1(threadfunc,1);
std::thread t2(threadfunc,2);
std::thread t3(threadfunc,3);
t1.join();
t2.join();
t3.join();
std::cout<<i<<std::endl;
}
This code will output "2349", "3249", "4239", "4329", "2439" or "3429", but never anything else. Each thread has its own copy of i, which is assigned to, incremented and then printed. The thread running main also has its own copy, which is assigned to at the beginning and then left unchanged. These copies are entirely independent, and each has a different address.
It is only the name that is special in that respect --- if you take the address of a thread_local variable then you just have a normal pointer to a normal object, which you can freely pass between threads. e.g.
thread_local int i=0;
void thread_func(int*p){
*p=42;
}
int main(){
i=9;
std::thread t(thread_func,&i);
t.join();
std::cout<<i<<std::endl;
}
Since the address of i is passed to the thread function, then the copy of i belonging to the main thread can be assigned to even though it is thread_local. This program will thus output "42". If you do this, then you need to take care that *p is not accessed after the thread it belongs to has exited, otherwise you get a dangling pointer and undefined behaviour just like any other case where the pointed-to object is destroyed.
thread_local variables are initialized "before first use", so if they are never touched by a given thread then they are not necessarily ever initialized. This is to allow compilers to avoid constructing every thread_local variable in the program for a thread that is entirely self-contained and doesn't touch any of them. e.g.
struct my_class{
my_class(){
std::cout<<"hello";
}
~my_class(){
std::cout<<"goodbye";
}
};
void f(){
thread_local my_class unused;
}
void do_nothing(){}
int main(){
std::thread t1(do_nothing);
t1.join();
}
In this program there are 2 threads: the main thread and the manually-created thread. Neither thread calls f, so the thread_local object is never used. It is therefore unspecified whether the compiler will construct 0, 1 or 2 instances of my_class, and the output may be "", "hellohellogoodbyegoodbye" or "hellogoodbye".
Thread-local storage is in every aspect like static (= global) storage, only that each thread has a separate copy of the object. The object's life time starts either at thread start (for global variables) or at first initialization (for block-local statics), and ends when the thread ends (i.e. when join() is called).
Consequently, only variables that could also be declared static may be declared as thread_local, i.e. global variables (more precisely: variables "at namespace scope"), static class members, and block-static variables (in which case static is implied).
As an example, suppose you have a thread pool and want to know how well your work load was being balanced:
thread_local Counter c;
void do_work()
{
c.increment();
// ...
}
int main()
{
std::thread t(do_work); // your thread-pool would go here
t.join();
}
This would print thread usage statistics, e.g. with an implementation like this:
struct Counter
{
unsigned int c = 0;
void increment() { ++c; }
~Counter()
{
std::cout << "Thread #" << std::this_thread::id() << " was called "
<< c << " times" << std::endl;
}
};
Let's say I create:
class Hello {
public:
int World(int in)
{
static int var = 0; // <<<< This thing here.
if (in >= 0) {
var = in;
} else {
cout << var << endl;
}
}
};
Now, if I do:
Hello A;
Hello B;
A.World(10);
A.World(-1);
B.World(-1);
I'm getting output of "10" followed by another "10". The value of the local variable of a method just crossed over from one instance of a class to another.
It's not surprising - technically methods are just functions with a hidden this parameter, so a static local variable should behave just like in common functions. But is it guaranteed? Is it a behavior enforced by standard, or is it merely a happy byproduct of how the compiler handles methods? In other words - is this behavior safe to use? (...beyond the standard risk of baffling someone unaccustomed...)
Yes. It doesn't matter if the function is a [non-static] member of a class or not, it's guranteed to have only one instance of it's static variables.
Proper technical explanation for such variables is that those are objects with static duration and internal linkage - and thus those names live until program exits, and all instances of this name refer to the same entity.
Just one thing to add to the correct answer. If your class was templated, then the instance of var would only be shared amongst objects of the same instantiation type. So if you had:
template<typename C>
class Hello {
public:
int World(int in)
{
static int var = 0; // <<<< This thing here.
if (in >= 0) {
var = in;
} else {
cout << var << endl;
}
}
};
And then:
Hello<int> A;
Hello<int> B;
Hello<unsigned> C;
A.World(10);
A.World(-1);
B.World(-1);
C.World(-1);
Then the final output would be "0" rather than "10", because the Hello<unsigned> instantiation would have its own copy of var.
If we are talking about the Windows Compiler it's guaranteed
https://msdn.microsoft.com/en-us/library/y5f6w579.aspx
The following example shows a local variable declared static in a member function. The static variable is available to the whole program; all instances of the type share the same copy of the static variable.
They use an example very similar to yours.
I don't know about GCC
Yes, it is guaranteed. Now, to answer the question "Any risk of sharing local static variable of a method between instances?" it might be a bit less straightforward. There might be potential risks in the initialization and utilization of the variable and these risks are specific to variables local to the method (as opposed to class variables).
For the initialization, a relevant part in the standard is 6.7/4 [stmt.dcl]:
Dynamic initialization of a block-scope variable with static storage
duration (3.7.1) or thread storage duration (3.7.2) is performed the
first time control passes through its declaration; such a variable is
considered initialized upon the completion of its initialization. If
the initialization exits by throwing an exception, the initialization
is not complete, so it will be tried again the next time control
enters the declaration. If control enters the declaration concurrently
while the variable is being initialized, the concurrent execution
shall wait for completion of the initialization. If control
re-enters the declaration recursively while the variable is being
initialized, the behavior is undefined.
In the simple cases, everything should work as expected. When the construction and initialization of the variable is more complex, there will be risks specific to this case. For instance, if the constructor throws, it will have the opportunity to throw again on the next call. Another example would be recursive initialization which is undefined behavior.
Another possible risk is the performance of the method. The compiler will need to implement a mechanism to ensure compliant initialization of the variable. This is implementation-dependent and it could very well be a lock to check if the variable is initialized, and that lock could be executed every time the method is called. When that happens, it can have a significant adverse effect on performance.
Recently my company has begun the process of upgrading to Visual Studio 2015 from Visual Studio 2010. The problem we're currently running into apparently seems to stem from a change in the behavior of the compiler. We can build and run our solution, but it seems to deadlock (it seems to just idle: CPU usage is nearly 0).
Stepping through with the debugger we've discovered an issue where a singleton object depends on itself during initialization. Here's an extremely stripped down version:
#include <iostream>
using namespace std;
struct Singleton
{
Singleton( int n )
{
cout << "Singleton( " << n << " )" << endl;
cout << Singleton::Instance().mN << endl;
mN = n;
}
static Singleton& Instance()
{
static Singleton instance( 5 );
return instance;
}
int mN;
};
int main() {
cout << Singleton::Instance().mN << endl;
return 0;
}
Naturally in our code there's a lot of other things going on, but this code exhibits the same behavior that we're seeing in the main project. In VS2010, this builds, runs, and terminates "normally". In VS2015 it deadlocks.
I've also tried this in ideone.com with various versions of C++ and all of those reproduce the deadlocking behavior. It makes sense to me that this doesn't work (nor should it work), because the object shouldn't depend on itself.
What I'm more curious about is why did this "work" in VS2010? What does the standard have to say about static variable initialization? Was this just a VS2010 (and possibly earlier) compiler bug?
The standard says that:
If control enters the declaration concurrently while the [block-scope variable with static or thread storage duration] is being initialized, the concurrent execution shall wait for completion of the initialization. If control re-enters the declaration recursively while the variable is being initialized, the behavior is undefined.
([stmt.dcl]/4)
The change made in C++11 is that initialization of local static variables is required to be thread-safe. The standard disallows recursion that would pass through the declaration again during the initialization, and the UB that results is manifesting as deadlock in your case---which makes perfect sense, since the second pass through the declaration is waiting forever for the first one to complete.
Now, this was undefined behavior in C++03 as well, but in a C++03 implementation, the initialization is not required to be thread-safe, so what probably happens is this: on the first pass through the declaration, a flag is set and then the constructor is called; the second pass sees the flag, assumes the variable is already initialized, and then returns a reference to it. Then the initialization completes.
You should rewrite your code, obviously, to avoid this recursive initialization.
I am confused with the description of thread_local in C++11. My understanding is, each thread has unique copy of local variables in a function. The global/static variables can be accessed by all the threads (possibly synchronized access using locks). And the thread_local variables are visible to all the threads but can only modified by the thread for which they are defined? Is it correct?
Thread-local storage duration is a term used to refer to data that is seemingly global or static storage duration (from the viewpoint of the functions using it) but, in actual fact, there is one copy per thread.
It adds to the current options:
automatic (exists during a block or function);
static (exists for the program duration); and
dynamic (exists on the heap between allocation and deallocation).
Something that is thread-local is brought into existence at thread creation time and disposed of when the thread finishes.
For example, think of a random number generator where the seed must be maintained on a per-thread basis. Using a thread-local seed means that each thread gets its own random number sequence, independent of all other threads.
If your seed was a local variable within the random function, it would be initialised every time you called it, giving you the same number each time. If it was a global, threads would interfere with each other's sequences.
Another example is something like strtok where the tokenisation state is stored on a thread-specific basis. That way, a single thread can be sure that other threads won't screw up its tokenisation efforts, while still being able to maintain state over multiple calls to strtok - this basically renders strtok_r (the thread-safe version) redundant.
Yet another example would be something like errno. You don't want separate threads modifying errno after one of your calls fails, but before you've had a chance to check the result.
This site has a reasonable description of the different storage duration specifiers.
When you declare a variable thread_local then each thread has its own copy. When you refer to it by name, then the copy associated with the current thread is used. e.g.
thread_local int i=0;
void f(int newval){
i=newval;
}
void g(){
std::cout<<i;
}
void threadfunc(int id){
f(id);
++i;
g();
}
int main(){
i=9;
std::thread t1(threadfunc,1);
std::thread t2(threadfunc,2);
std::thread t3(threadfunc,3);
t1.join();
t2.join();
t3.join();
std::cout<<i<<std::endl;
}
This code will output "2349", "3249", "4239", "4329", "2439" or "3429", but never anything else. Each thread has its own copy of i, which is assigned to, incremented and then printed. The thread running main also has its own copy, which is assigned to at the beginning and then left unchanged. These copies are entirely independent, and each has a different address.
It is only the name that is special in that respect --- if you take the address of a thread_local variable then you just have a normal pointer to a normal object, which you can freely pass between threads. e.g.
thread_local int i=0;
void thread_func(int*p){
*p=42;
}
int main(){
i=9;
std::thread t(thread_func,&i);
t.join();
std::cout<<i<<std::endl;
}
Since the address of i is passed to the thread function, then the copy of i belonging to the main thread can be assigned to even though it is thread_local. This program will thus output "42". If you do this, then you need to take care that *p is not accessed after the thread it belongs to has exited, otherwise you get a dangling pointer and undefined behaviour just like any other case where the pointed-to object is destroyed.
thread_local variables are initialized "before first use", so if they are never touched by a given thread then they are not necessarily ever initialized. This is to allow compilers to avoid constructing every thread_local variable in the program for a thread that is entirely self-contained and doesn't touch any of them. e.g.
struct my_class{
my_class(){
std::cout<<"hello";
}
~my_class(){
std::cout<<"goodbye";
}
};
void f(){
thread_local my_class unused;
}
void do_nothing(){}
int main(){
std::thread t1(do_nothing);
t1.join();
}
In this program there are 2 threads: the main thread and the manually-created thread. Neither thread calls f, so the thread_local object is never used. It is therefore unspecified whether the compiler will construct 0, 1 or 2 instances of my_class, and the output may be "", "hellohellogoodbyegoodbye" or "hellogoodbye".
Thread-local storage is in every aspect like static (= global) storage, only that each thread has a separate copy of the object. The object's life time starts either at thread start (for global variables) or at first initialization (for block-local statics), and ends when the thread ends (i.e. when join() is called).
Consequently, only variables that could also be declared static may be declared as thread_local, i.e. global variables (more precisely: variables "at namespace scope"), static class members, and block-static variables (in which case static is implied).
As an example, suppose you have a thread pool and want to know how well your work load was being balanced:
thread_local Counter c;
void do_work()
{
c.increment();
// ...
}
int main()
{
std::thread t(do_work); // your thread-pool would go here
t.join();
}
This would print thread usage statistics, e.g. with an implementation like this:
struct Counter
{
unsigned int c = 0;
void increment() { ++c; }
~Counter()
{
std::cout << "Thread #" << std::this_thread::id() << " was called "
<< c << " times" << std::endl;
}
};
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