I'm trying to implement timer with standard environment
Here is a code I have:
bool shutdownDetected = false;
void signal_handler(const int sigid)
{
shutdownDetected = true;
}
int main(int argc, const char * argv[])
{
signal(SIGTERM, (sig_t)signal_handler);
std::async(std::launch::async, [&] () {
std::this_thread::sleep_for( std::chrono::milliseconds{5000});
std::cout << "On TIMER!" << std::endl;
} );
std::cout << "main function" << std::endl;
while (!shutdownDetected) {
}
return EXIT_SUCCESS;
}
As result I see in output after 5 seconds:
// 5 seconds left
On Timer
main function
but would like to see:
main function
// 5 seconds left
On Timer
Seems that my implementation hangs main thread as well. How to avoid this?
Your std::async command returns an std::future, which is then immediately destroyed. The problem is that destruction of a future involves 'joining' the thread you created, which means that the destructor is going to wait until the thread has ended itself and code execution in your main thread doesn't advance until that process has completed.
Simple answer is to assign the result of your std::async call to a variable, and possibly call its get() member function in your loop that tests for termination.
auto t = std::async(std::launch::async, [&] () {
std::this_thread::sleep_for( std::chrono::milliseconds{5000});
std::cout << "On TIMER!" << std::endl;
} );
std::cout << "main function" << std::endl;
t.get();
std::async(std::launch::async, [&] () {
std::this_thread::sleep_for( std::chrono::milliseconds{5000});
std::cout << "On TIMER!" << std::endl;
} );
Does not work unless you assign the std::future returned by std::async to a variable and keep it around. I did not know why this is, clearly because I couldn't be bothered to look it up. Vincent Savard did, and linked us to documentation on the destructor for std::future which says:
it may block if all of the following are true: the shared state was created by a call to std::async, the shared state is not yet ready, and this was the last reference to the shared state.
Since the returnded std::future is not assigned to anything, it is instantly destroyed and the destructor blocks until completion.
I'm going to leave out the signal handler as it's not relevant to the problem.
#include <iostream>
#include <future>
int main()
{
auto letMeLive = std::async(std::launch::async, [] () {
std::this_thread::sleep_for( std::chrono::milliseconds{5000});
std::cout << "On TIMER!" << std::endl;
} );
std::cout << "main function" << std::endl;
letMeLive.wait(); // instead of the signal handler
return EXIT_SUCCESS;
}
Related
I am starting a std::async from within a lambda function.
Even with the policy std::launch::async, the task is running synchronously on the same thread and therefore blocking it.
Is this normal or am I missing something?
int main()
{
auto lambda = [&]
{
auto future = std::async(std::launch::async, [&]
{
using namespace std::chrono_literals;
const auto delay = 5000ms;
std::this_thread::sleep_for(delay);
std::cout << "Done okay byeeeeeeee \n";
});
};
lambda();
for (long long i = 0 ; i < 10 ; ++i)
{
std::cout << "Doing stuff in main thread" << std::endl;
}
}
The problem is the future variable, or rather the destruction of it.
The destructor can block until the async is finished.
I had a comparable problem last and didn't solve it with std::async but with std::thread.
void get_sleep()
{
std::this_thread::sleep_for(std::chrono::seconds(5));
std::cout << "Done okay byeeeeeeee \n";
}
int main()
{
std::thread t(get_sleep);
t.detach();
for (int i = 0 ; i < 10 ; ++i)
{
std::cout << "Doing stuff in main thread" << std::endl;
}
return 0;
}
Expanding upon Some programmer dude's answer
Consider the following class→ I created this to see when the Destructor gets called.
struct MyStruct {
~MyStruct() {printf("DTOR\r\n");}
};
Here is your code, I have heavily commented the code, so let the code do the talking
int main(){
//1 You are on main thread, create lambda on main thread
auto lambda = [&]{
//3 lambda is being started to execute
MyStruct myStruct{}; //4 This I added
//4 std::async enqueues the inner lambda, which will run and return in the future
auto future = std::async(std::launch::async, [&]
{
using namespace std::chrono_literals;
const auto delay = 5000ms;
std::this_thread::sleep_for(delay);
std::cout << "Done okay byeeeeeeee \n";
});
//5 Going out of scope, destructors will be called therefore you will be blocked
};
//2 You are on main thread
// You execute lambda on main thread, now you will enter into its contents
lambda();
//6 After all destructors run you will continue here
I am trying to run run() function every 5 seconds without stopping while() loop (parallelly). How can I do that ? Thanks in advance
#include <iostream>
#include <thread>
#include <chrono>
using namespace std;
void run()
{
this_thread::sleep_for(chrono::milliseconds(5000));
cout << "good morning" << endl;
}
int main()
{
thread t1(run);
t1.detach();
while(1)
{
cout << "hello" << endl;
this_thread::sleep_for(chrono::milliseconds(500));
}
return 0;
}
In your main function, it is important to understand what each thread is doing.
The main thread creates a std::thread called t1
The main thread continues and detaches the thread
The main thread executes your while loop in which it:
prints hello
sleeps for 0.5 seconds
The main thread returns 0, your program is finished.
Any time from point 1, thread t1 sleeps for 5 seconds and then prints good morning. This happens only once! Also, as pointed out by #Fareanor, std::cout is not thread-safe, so accessing it with the main thread and thread t1 may result in a data race.
When the main thread reaches point 4 (it actually never does because your while loop is infinite), your thread t1 might have finished it's task or not. Imagine the potential problems that could occur. In most of the cases, you'll want to use std::thread::join().
To solve your problem, there are several alternatives. In the following, we will assume that the execution of the function run without the std::this_thread::sleep_for is insignificant compared to 5 seconds, as per the comment of #Landstalker. The execution time of run will then be 5 seconds plus some insignificant time.
As suggested in the comments, instead of executing the function run every 5 seconds, you could simply execute the body of run every 5 seconds by placing a while loop inside of that function:
void run()
{
while (true)
{
std::this_thread::sleep_for(std::chrono::milliseconds(5000));
std::cout << "good morning" << std::endl;
}
}
int main()
{
std::thread t(run);
t.join();
return 0;
}
If, for some reason, you really need to execute the run function every 5 seconds as stated in your question, you could launch a wrapper function or lambda which contains the while loop:
void run()
{
std::this_thread::sleep_for(std::chrono::milliseconds(5000));
std::cout << "good morning" << std::endl;
}
int main()
{
auto exec_run = [](){ while (true) run(); };
std::thread t(exec_run);
t.join();
return 0;
}
As a side note, it's better to avoid using namespace std.
Just call your run function in seperate thread function like below. Is this ok for you?
void ThreadFunction()
{
while(true) {
run();
this_thread::sleep_for(chrono::milliseconds(5000));
}
}
void run()
{
cout << "good morning" << endl;
}
int main()
{
thread t1(ThreadFunction);
t1.detach();
while(1)
{
cout << "hello" << endl;
this_thread::sleep_for(chrono::milliseconds(500));
}
return 0;
}
Sometimes it would be useful if a joinable std::thread had the hability to execute thread::join() on its destructor. See the examples below.
Example 1 (error):
The object std::thread has been destroyed after the throw of the exception. Once the the flow exits the scope, the destructor is called BEFORE the join happens. It makes STL show an error message of 'abort'.
int main( int argc, const char * argv[] )
{
try
{
thread t( [] ()
{
this_thread::sleep_for( chrono::seconds( 1 ) );
cout << "thread done" << endl;
} );
throw exception( "some exception" );
t.join();
}
catch ( const exception & )
{
cout << "exception caught!" << endl;
}
cout << "main done" << endl;
return 0;
}
Example 2 (correct way):
The object t is created before my try-catch block and the join() is put on both try and catch blocks. So it guarantees that the join() happens.
int main( int argc, const char * argv[] )
{
thread t;
try
{
t = thread( [] ()
{
this_thread::sleep_for( chrono::seconds( 1 ) );
cout << "thread done" << endl;
} );
throw exception( "some exception" );
t.join( );
}
catch ( const exception & )
{
t.join();
cout << "exception caught!" << endl;
}
cout << "main done" << endl;
return 0;
}
...AND THE QUESTION IS:
What is the reason for a joinable std::thread not join automatically on its destructor?
It would be much easier if it happened automatically. The way it's done today requires one must be careful when using threads inside try-catch blocks, for example... but I am sure someone THOUGHT when designed std::thread this way. So there must be a reason for that... what is that reason?
PS: I know we can envolve std::thread in a class and put the join() on the destructor of this new class... so it becomes automatic. But this is not the point. My question is really about std::thread itself.
The reason is simply so that you are forced to think about it. If a std::thread object is destroyed due to an exception escaping the scope then a join may cause a blocking wait during stack unwinding, which is often undesirable, and can lead to deadlock if the thread that is being waited for is in turn waiting for some action on the part of the thread doing the waiting.
By having the application terminate in this situation you as a programmer are forced to actively think about the conditions that would cause the object to be destroyed, and ensure that the thread is joined correctly.
I am running Visual Studio 2012 and attempting to learn how std::async works. I have created a very simple C++ console application:
#include "stdafx.h"
#include <future>
#include <iostream>
void foo() {
std::cout << "foo() thread sleep" << std::endl;
std::this_thread::sleep_for(std::chrono::seconds(5));
std::cout << "foo() thread awake" << std::endl;
}
int main()
{
std::future<void> res = std::async(std::launch::async, foo);
res.get();
std::cout << "MAIN THREAD" << std::endl;
system("pause");
return 0;
}
My initial expectation was to see "MAIN THREAD" printout appearing before "foo() thread awake" since the two threads are running asynchronously, with the foo() trailing behind due to its sleeping behavior. However, that is not what is actually happening. The call to res.get() blocks until foo() wakes up, and only then does it get to the "MAIN THREAD" printout. This is indicative of a synchronous behavior, so I am wondering what if perhaps I am either missing something, or not fully grasping the implementation. I have looked through numerous posts on this matter, but still cannot make any sense of it. Any help would be appreciated!
res.get();
blocks until the async is done.
http://en.cppreference.com/w/cpp/thread/future/get
Regardless of how you tell it to run, get can't give you the results until it's done.
Well, this is how std::future::get works - it blocks until future has some result or exception to provide.
that doesn't mean that async works synchronously, it is working asynchronously, it's only because you block the thread which waits on the result.
the idea was to to launch some task asynchronously, do something meanwhile and only call get when you need the result, as you might figured out, it is not the most scale-able thing..
if you use Visual Studio 2015, you can access the await keyword both for std::future and concurrency::task (Microsoft PPL library) , and for your own compatible defined types. this achieves non-blocking behavior.
#include "stdafx.h"
#include <future>
#include <iostream>
void foo() {
std::cout << "foo() thread sleep" << std::endl;
std::this_thread::sleep_for(std::chrono::seconds(5));
std::cout << "foo() thread awake" << std::endl;
}
std::future<void> entry(){
await std::async(std::launch::async, foo);
std::cout << "foo has finished, back in entry()\n";
}
int main()
{
std::cout << "MAIN THREAD" << std::endl;
entry();
std::cout << "BACK INMAIN THREAD" << std::endl;
system("pause");
return 0;
} //make sure to compile with /await flag
The problem is that res.get() has to wait for its thread to finish before getting its result (if any). To see the concurrency in motion you need to move the get() to after the other code that you want to run at the same time.
This example may make it a little clearer:
#include <ctime>
#include <cstdlib>
#include <future>
#include <iostream>
void foo(int id) {
std::cout << "foo(" << id << ") thread sleep" << std::endl;
// random sleep
std::this_thread::sleep_for(std::chrono::seconds(std::rand() % 10));
std::cout << "foo(" << id << ") thread awake" << std::endl;
}
int main()
{
std::srand(std::time(0));
std::future<void> res1 = std::async(std::launch::async, foo, 1);
std::future<void> res2 = std::async(std::launch::async, foo, 2);
std::future<void> res3 = std::async(std::launch::async, foo, 3);
std::cout << "MAIN THREAD SLEEPING" << std::endl;
std::this_thread::sleep_for(std::chrono::seconds(20));
std::cout << "MAIN THREAD AWAKE" << std::endl;
// now wait for all the threads to end
res1.get();
res2.get();
res3.get();
system("pause");
return 0;
}
I am beginning to use the thread class.
In the main() thread below, an Example class is created.
Inside the constructor of Example, two threads are created in the Example::start() function.
Example::foo() is designed to print a message every second.
Example::bar() is designed to print a message every 5 seconds.
Inside the main() function, a loop is designed to print every 3 seconds.
I decided to not use join() in Example::start() because I would like to have the main() function continuously run.
Why does the main thread crash during run-time?
// thread example
#include <iostream> // std::cout
#include <thread> // std::thread
#include <chrono> // std::chrono::seconds
using namespace std;
class Example
{
public:
Example();
void start();
void foo();
void bar(int x);
};
Example::Example()
{
start();
}
void Example::start()
{
std::thread first (&Example::foo, this); // spawn new thread that calls foo()
std::thread second (&Example::bar, this, 5); // spawn new thread that calls bar(0)
// synchronize threads:
//first.join(); // pauses until first finishes
//second.join(); // pauses until second finishes
}
void Example::foo()
{
cout << "entered foo()" << endl;
int count = 0;
while(1) {
std::this_thread::sleep_for (std::chrono::seconds(1));
++count;
cout << "foo() count = " << count << endl;
}
}
void Example::bar(int x)
{
cout << "entered bar() x = " << x << endl;
int count = 0;
while(1) {
std::this_thread::sleep_for (std::chrono::seconds(5));
++count;
cout << "bar() count = " << count << endl;
}
}
int main() {
Example* c = new Example();
cout << "Example() created" << endl;
while(true) {
std::this_thread::sleep_for(std::chrono::seconds(3));
cout << "main() thread loop..." << endl;
}
std::cout << "end of main()";
delete c;
return 0;
}
Foo::Start() initalizes two threads, thread Foo and thread bar. When the function Start returns to the main thread, the two thread objects go out of scope and the destructor is called for clearing out of scope variables.
A simple solution would be to make threads part of the class.
On another note, std::cout is not a synchronized class, when writing your text might be garbled: Is cout synchronized/thread-safe?
Also, when creating your class Example, delete is never called which causes a memory leak.
Your comment here:
void Example::start()
{
std::thread first (&Example::foo, this); // spawn new thread that calls foo()
std::thread second (&Example::bar, this, 5); // spawn new thread that calls bar(0)
// synchronize threads:
//first.join(); // pauses until first finishes
//second.join(); // pauses until second finishes
}
Is wrong.
Not only does the the join pause until the threads finish. But they also allow the thread to be cleaned up. A thread destructor calls terminate while the thread is join-able (ie it is still running). So you must call join() on the thread (to wait for it to finish) before you can allow the destructor to be called.
One of the comments above suggests calling detach(). This detaches the thread of execution from the thread object (thus making it not join-able). This will work (as your code is in infinite loop), but is a bad idea generally. As allowing main() to exit while threads are still running is undefined behavior.