In C++, I have this sample implementation:
#include <thread>
#include <iostream>
void doSomeWork( void )
{
std::cout << "hello from thread..." << std::endl;
while(true)
{
printf("Getting inside doSomeWork\n");
sleep(1);
}
}
int main( int argc, char *argv[] )
{
std::thread t( doSomeWork );
t.join();
processing();
return 0;
}
void processing()
{
while(true)
{
printf("Getting inside processing\n");
sleep(1);
}
}
I have a problem that doSomeWork() keep doing things and it block processing(). I thought thread is asynchronous so that while it is sleeping, I can other things. My question here is how can I sleep in doSomeWork() and yield other threads, then resume doSomework()?
The line
t.join();
pauses the calling thread until thread t completes (when doSomeWork()) returns.
Change your code to
std::thread t( doSomeWork );
processing();
t.join();
if you want processing() and doSomeWork() to run in parallel.
replace t.join(); with t.detach();
Note however, detach means something like "I'm not interesting in this thread, so it may be terminated in any time" or "I will synchronize the thread by some sync mechanism". Basically it is used when you have some external(to the thread) sync mechanism so you don't want to use join. It is common to use join every time but use detach only when needed. In your particular case you have sync mechanism with forever loop, so it is absolutely fine to use detach but be aware about their differences.
Related
What would happen after running these code below?
#include <thread>
#include <chrono>
#include <functional>
#include <fstream>
using namespace std;
void func()
{
std::this_thread::sleep_for(std::chrono::seconds(5));
ofstream outfile("test.txt");
outfile << "hello world" << endl;
outfile.close();
}
void start()
{
std::thread th(std::bind(func));
if (th.joinable())
th.detach();
}
int main()
{
start();
return 0;
}
Result is that "test.txt" file will not be created in the disk. why?
In addition, are there problems if I use the heap data which new in start() function in the func()? The os will delete it when the main thread return but the child thread still running?
As #Brandon has pointed out in the comments, your thread very likely doesn't get a chance to run. main() calls start() which constructs the thread, then returns back to main() which then exits. Even if your thread got a chance to run in that time, you put a 5s sleep in there.
You need to wait for your thread to finish before terminating the program. You can use std::thread::join for this:
void start()
{
std::thread th(std::bind(func));
// if the thread was successfully created
if (th.joinable())
// wait for it to finish
th.join();
}
You could also consider std::condition_variable if you didn't need to wait entirely on your thread, and your thread could alert you when some of its work is done.
So I have this class:
class foo {
public:
foo() { };
void me1() const {
while(1) {
std::lock_guard<std::mutex> ldock(m);
std::cout << 0;
}
}
void me2() const {
while(1) {
std::lock_guard<std::mutex> ldock(m);
std::cout << 1;
}
}
private:
std::mutex m;
};
Now I want to run this two methods in some two different threads, I do it like this:
int main() {
foo myfoo;
std::thread firstThread(&foo::me1, &myfoo);
std::thread secondThread(&foo::me2, &myfoo);
firstThread.detach();
secondThread.detach();
//while(1) { }
return 0;
}
I don't want to wait for any of this two methods to finish, they will simultaneously run until the main thread will be killed.
Is it ok to have some kind of infinite-loop at the end of main thread? (like the commented while(1) {}).
Or should I call some kinda sleep function?
You need to define an exit condition in your foo::me1() and foo::me2() . If you don't know how to do that, that
sleep(/*number of seconds you want your program to run*/ );
will do just fine.
If you define a termination clause then the bruteforce would be
to expose something like an atomic:
class foo {
public:
std::atomic<int> me1done = false;
std::atomic<int> me2done = false;
foo() { };
void me1() {
while(/* need exit condition here*/) {
std::lock_guard<std::mutex> ldock(m);
std::cout << 0;
}
me1done = true;
}
void me2() {
while(/*need exit condition here*/) {
std::lock_guard<std::mutex> ldock(m);
std::cout << 1;
}
me2done = true;
}
private:
std::mutex m;
};
and then you can check in main by polling every x-seconds.
int main(void)
{
// start your threads and detach
foo myfoo;
std::thread firstThread(&foo::me1, &myfoo);
std::thread secondThread(&foo::me2, &myfoo);
firstThread.detach();
secondThread.detach();
while( not (myfoo.me1done and myfoo.me2done ) )
{
sleep( /* some time */);
}
return 0;
}
If you want to be more elaborate you will have to work with condition variables.
If you want to determine if the two threads have finished your best bet is actually not to detach() the threads but rather join() them before exiting the main thread. That is, you'd kick off both threads and they'll run concurrently and once kicked off you simply join() each. Of course, that assumes that the threads would terminate.
Having a detach()ed thread effectively means you can never be sure if it has finished. That is generally rarely useful and I consider it a mistake that detach() was added to std::thread. However, even with detach()ed thread you can recognize when an objective is achieved without a busy wait. To that end you'd set up suitable variables indicating completion or progress and have them protected by a std::mutex. The main thread would then wait() on a std::condition_variable which gets notify_once()ed by the respective thread upon the completion/progress update which would be done in reasonable intervals. Once all threads have indicated that they are done or have achieved a suitable objective the main() thread can finish.
Using a timer alone is generally not a good approach. The signalling between threads is typically preferable and tends to create a more responsive system. You can still used a timed version of wait() (i.e., wait_until() or wait_for()), e.g., to alert upon suspecting a somehow hung or timed-out thread.
empty infinite loops as while(1) { } are UB.
adding a sleep inside is OK though.
To run infinitely foo::me1/foo::me2, you have several other choices:
int main()
{
foo myfoo;
std::thread firstThread(&foo::me1, &myfoo);
std::thread secondThread(&foo::me2, &myfoo);
firstThread.join(); // wait infinitely as it never ends.
secondThread.join(); // and so never reach
}
or simply use main thread to do one work:
int main()
{
foo myfoo;
std::thread firstThread(&foo::me1, &myfoo);
myfoo.me2(); // work infinitely as it never ends.
firstThread.join(); // and so never reach
}
Can B thread can created in A thread?
After waiting for B thread end, Can A thread continue to run?
Short answer
Yes
Yes
There is very little conceptual difference between thread A and the main thread. Note that you could even join thread B in the main thread even though it was created from thread A.
Sample: (replace <thread> with <boost/thread.hpp> if you don't have a c++11 compiler yet)
Live On Coliru
#include <thread>
#include <iostream>
void threadB() {
std::cout << "Hello world\n";
}
void threadA() {
std::thread B(threadB);
B.join();
std::cout << "Continued to run\n";
}
int main() {
std::thread A(threadA);
A.join(); // no difference really
}
Prints
Hello world
Continued to run
If B is a child thread of A?
There are ways to synchronize threads for turn taking. Whether or not they can run in parallel depends on using kernel threads or user threads. User threads are not aware of different processors so they cannot run truly in 'parallel'. If you want the threads to take turns you can use a mutex/semaphore/lock to synchronize them. If you want them to run in true parallel you will need B to be a child process of A.
You can also end the child thread/process in which case the parent will be scheduled. It's often not possible to guarantee scheduling without some sort of synchronization.
void FuncA()
{
if(ScanResultsMonitorThread == NULL) {
/* start thread A */
}
}
void FunAThread()
{
while(1) {
FuncB();
}
}
void FuncB()
{
try {
boost::this_thread::sleep(boost::posix_time::seconds(25));
}
catch(const boost::thread_interrupted&) {
}
if(needRestart){
/* create thread B */
boost::thread Restart(&FuncBThread,this);
boost::this_thread::sleep(boost::posix_time::seconds(10));
/* program can not run here and thread A end, why? */
}
else {
}
}
I have a vector of Timer Objects. Each Timer Object launches an std::thread that simulates a growing period. I am using a Command pattern.
What is happening is each Timer is getting executed one after another but what I really want is for one to be executed....then once finished, the next one...once finished the next...while not interfering with the main execution of the program
class Timer
{
public:
bool _bTimerStarted;
bool _bTimerCompleted;
int _timerDuration;
virtual ~Timer() { }
virtual void execute()=0;
virtual void runTimer()=0;
inline void setDuration(int _s) { _timerDuration = _s; };
inline int getDuration() { return _timerDuration; };
inline bool isTimerComplete() { return _bTimerCompleted; };
};
class GrowingTimer : public Timer
{
public:
void execute()
{
//std::cout << "Timer execute..." << std::endl;
_bTimerStarted = false;
_bTimerCompleted = false;
//std::thread t1(&GrowingTimer::runTimer, this); //Launch a thread
//t1.detach();
runTimer();
}
void runTimer()
{
//std::cout << "Timer runTimer..." << std::endl;
_bTimerStarted = true;
auto start = std::chrono::high_resolution_clock::now();
std::this_thread::sleep_until(start + std::chrono::seconds(20));
_bTimerCompleted = true;
std::cout << "Growing Timer Finished..." << std::endl;
}
};
class Timers
{
std::vector<Timer*> _timers;
struct ExecuteTimer
{
void operator()(Timer* _timer) { _timer->execute(); }
};
public:
void add_timer(Timer& _timer) { _timers.push_back(&_timer); }
void execute()
{
//std::for_each(_timers.begin(), _timers.end(), ExecuteTimer());
for (int i=0; i < _timers.size(); i++)
{
Timer* _t = _timers.at(i);
_t->execute();
//while ( ! _t->isTimerComplete())
//{
//}
}
}
};
Executing the above like:
Timers _timer;
GrowingTimer _g, g1;
_g.setDuration(BROCCOLI::growTimeSeconds);
_g1.setDuration(BROCCOLI::growTimeSeconds);
_timer.add_timer(_g);
_timer.add_timer(_g1);
start_timers();
}
void start_timers()
{
_timer.execute();
}
In Timers::execute I am trying a few different ways to execute the first and not execute the
next until I somehow signal it is done.
UPDATE:
I am now doing this to execute everything:
Timers _timer;
GrowingTimer _g, g1;
_g.setDuration(BROCCOLI::growTimeSeconds);
_g1.setDuration(BROCCOLI::growTimeSeconds);
_timer.add_timer(_g);
_timer.add_timer(_g1);
//start_timers();
std::thread t1(&Broccoli::start_timers, this); //Launch a thread
t1.detach();
}
void start_timers()
{
_timer.execute();
}
The first time completes (I see the "completed" cout), but crashes at _t->execute(); inside the for loop with an EXEC_BAD_ACCESS. I added a cout to check the size of the vector and it is 2 so both timers are inside. I do see this in the console:
this Timers * 0xbfffd998
_timers std::__1::vector<Timer *, std::__1::allocator<Timer *> >
if I change the detach() to join() everything completes without the crash, but it blocks execution of my app until those timers finish.
Why are you using threads here? Timers::execute() calls execute on a timer, then waits for it to finish, then calls execute on the next, and so forth. Why don't you just call the timer function directly in Timers::execute() rather than spawning a thread and then waiting for it?
Threads allow you to write code that executes concurrently. What you want is serial execution, so threads are the wrong tool.
Update: In the updated code you run start_timers on a background thread, which is good. However, by detaching that thread you leave the thread running past the end of the scope. This means that the timer objects _g and _g1 and even the Timers object _timers are potentially destroyed before the thread has completed. Given the time-consuming nature of the timers thread, and the fact that you used detach rather than join in order to avoid your code blocking, this is certainly the cause of your problem.
If you run code on a thread then you need to ensure that all objects accessed by that thread have a long-enough lifetime that they are still valid when the thread accesses them. For detached threads this is especially hard to achieve, so detached threads are not recommended.
One option is to create an object containing _timers, _g and _g1 along side the thread t1, and have its destructor join with the thread. All you need to do then is to ensure that the object lives until the point that it is safe to wait for the timers to complete.
If you don't want to interfere with the execution of the program, you could do something like #Joel said but also adding a thread in the Timers class which would execute the threads in the vector.
You could include a unique_ptr to the thread in GrowingTimer instead of creating it as a local object in execute and calling detach. You can still create the thread in execute, but you would do it with a unique_ptr::reset call.
Then use join instead of isTimerComplete (add a join function to the Timer base class). The isTimerComplete polling mechanism will be extremely inefficient because it will basically use up that thread's entire time slice continually polling, whereas join will block until the other thread is complete.
An example of join:
#include <iostream>
#include <chrono>
#include <thread>
using namespace std;
void threadMain()
{
this_thread::sleep_for(chrono::seconds(5));
cout << "Done sleeping\n";
}
int main()
{
thread t(threadMain);
for (int i = 0; i < 10; ++i)
{
cout << i << "\n";
}
t.join();
cout << "Press Enter to exit\n";
cin.get();
return 0;
}
Note how the main thread keeps running while the other thread does its thing. Note that Anthony's answer is right in that it doesn't really seem like you need more than one background thread that just executes tasks sequentially rather than starting a thread and waiting for it to finish before starting a new one.
IS there a way of running a function back on the main thread ?
So if I called a function via Async that downloaded a file and then parsed the data. It would then call a callback function which would run on my main UI thread and update the UI ?
I know threads are equal in the default C++ implementation so would I have to create a shared pointer to my main thread. How would I do this and pass the Async function not only the shared pointer to the main thread but also a pointer to the function I want to rrun on it and then run it on that main thread ?
I have been reading C++ Concurrency in Action and chapter four (AKA "The Chapter I Just Finished") describes a solution.
The Short Version
Have a shared std::deque<std::packaged_task<void()>> (or a similar sort of message/task queue). Your std::async-launched functions can push tasks to the queue, and your GUI thread can process them during its loop.
There Isn't Really a Long Version, but Here Is an Example
Shared Data
std::deque<std::packaged_task<void()>> tasks;
std::mutex tasks_mutex;
std::atomic<bool> gui_running;
The std::async Function
void one_off()
{
std::packaged_task<void()> task(FUNCTION TO RUN ON GUI THREAD); //!!
std::future<void> result = task.get_future();
{
std::lock_guard<std::mutex> lock(tasks_mutex);
tasks.push_back(std::move(task));
}
// wait on result
result.get();
}
The GUI Thread
void gui_thread()
{
while (gui_running) {
// process messages
{
std::unique_lock<std::mutex> lock(tasks_mutex);
while (!tasks.empty()) {
auto task(std::move(tasks.front()));
tasks.pop_front();
// unlock during the task
lock.unlock();
task();
lock.lock();
}
}
// "do gui work"
std::this_thread::sleep_for(std::chrono::milliseconds(1000));
}
}
Notes:
I am (always) learning, so there is a decent chance that my code is not great. The concept is at least sound though.
The destructor of the return value from std::async (a std::future<>) will block until the operation launched with std::async completes (see std::async ), so waiting on the result of a task (as I do in my example) in one_off might not be a brilliant idea.
You may want to (I would, at least) create your own threadsafe MessageQueue type to improve code readability/maintainability/blah blah blah.
I swear there was one more thing I wanted to point out, but it escapes me right now.
Full Example
#include <atomic>
#include <chrono>
#include <deque>
#include <iostream>
#include <mutex>
#include <future>
#include <thread>
// shared stuff:
std::deque<std::packaged_task<void()>> tasks;
std::mutex tasks_mutex;
std::atomic<bool> gui_running;
void message()
{
std::cout << std::this_thread::get_id() << std::endl;
}
void one_off()
{
std::packaged_task<void()> task(message);
std::future<void> result = task.get_future();
{
std::lock_guard<std::mutex> lock(tasks_mutex);
tasks.push_back(std::move(task));
}
// wait on result
result.get();
}
void gui_thread()
{
std::cout << "gui thread: "; message();
while (gui_running) {
// process messages
{
std::unique_lock<std::mutex> lock(tasks_mutex);
while (!tasks.empty()) {
auto task(std::move(tasks.front()));
tasks.pop_front();
// unlock during the task
lock.unlock();
task();
lock.lock();
}
}
// "do gui work"
std::this_thread::sleep_for(std::chrono::milliseconds(1000));
}
}
int main()
{
gui_running = true;
std::cout << "main thread: "; message();
std::thread gt(gui_thread);
for (unsigned i = 0; i < 5; ++i) {
// note:
// these will be launched sequentially because result's
// destructor will block until one_off completes
auto result = std::async(std::launch::async, one_off);
// maybe do something with result if it is not void
}
// the for loop will not complete until all the tasks have been
// processed by gui_thread
// ...
// cleanup
gui_running = false;
gt.join();
}
Dat Output
$ ./messages
main thread: 140299226687296
gui thread: 140299210073856
140299210073856
140299210073856
140299210073856
140299210073856
140299210073856
Are you looking for std::launch::deferred ? Passing this parameter to std::async makes the task executed on the calling thread when the get() function is called for the first time.