JAVA THREAD POOL | SDRUŽOVÁNÍ VLÁKEN V JAVĚ

Java Thread pool představuje skupinu pracovních vláken, která čekají na úlohu a jsou mnohokrát znovu použita.

V případě fondu vláken se vytvoří skupina vláken s pevnou velikostí. Vlákno z fondu vláken je vytaženo a poskytovatel služeb mu přiřadí úlohu. Po dokončení úlohy je vlákno znovu obsaženo ve fondu vláken.

Metody fondu závitů

newFixedThreadPool(int s): Metoda vytvoří fond vláken s pevnou velikostí s.

newCachedThreadPool(): Metoda vytvoří nový fond vláken, který v případě potřeby vytvoří nová vlákna, ale stále bude používat dříve vytvořené vlákno, kdykoli budou k dispozici k použití.

newSingleThreadExecutor(): Metoda vytvoří nové vlákno.

Výhoda Java Thread Pool

Lepší výkon Šetří to čas, protože není potřeba zakládat nové vlákno.

Využití v reálném čase

Používá se v Servletu a JSP, kde kontejner vytvoří fond vláken pro zpracování požadavku.

Příklad Java Thread Pool

Podívejme se na jednoduchý příklad fondu vláken Java pomocí ExecutorService a Executors.

Soubor: WorkerThread.java

 import java.util.concurrent.ExecutorService; import java.util.concurrent.Executors; class WorkerThread implements Runnable { private String message; public WorkerThread(String s){ this.message=s; } public void run() { System.out.println(Thread.currentThread().getName()+&apos; (Start) message = &apos;+message); processmessage();//call processmessage method that sleeps the thread for 2 seconds System.out.println(Thread.currentThread().getName()+&apos; (End)&apos;);//prints thread name } private void processmessage() { try { Thread.sleep(2000); } catch (InterruptedException e) { e.printStackTrace(); } } }

Soubor: TestThreadPool.java

 public class TestThreadPool { public static void main(String[] args) { ExecutorService executor = Executors.newFixedThreadPool(5);//creating a pool of 5 threads for (int i = 0; i <10; i++) { runnable worker="new" workerthread('' + i); executor.execute(worker); calling execute method of executorservice } executor.shutdown(); while (!executor.isterminated()) system.out.println('finished all threads'); < pre> <p> <strong>Output:</strong> </p> <pre>pool-1-thread-1 (Start) message = 0 pool-1-thread-2 (Start) message = 1 pool-1-thread-3 (Start) message = 2 pool-1-thread-5 (Start) message = 4 pool-1-thread-4 (Start) message = 3 pool-1-thread-2 (End) pool-1-thread-2 (Start) message = 5 pool-1-thread-1 (End) pool-1-thread-1 (Start) message = 6 pool-1-thread-3 (End) pool-1-thread-3 (Start) message = 7 pool-1-thread-4 (End) pool-1-thread-4 (Start) message = 8 pool-1-thread-5 (End) pool-1-thread-5 (Start) message = 9 pool-1-thread-2 (End) pool-1-thread-1 (End) pool-1-thread-4 (End) pool-1-thread-3 (End) pool-1-thread-5 (End) Finished all threads </pre> download this example <h2>Thread Pool Example: 2</h2> <p>Let&apos;s see another example of the thread pool.</p> <p> <strong>FileName:</strong> ThreadPoolExample.java</p> <pre> // important import statements import java.util.Date; import java.util.concurrent.ExecutorService; import java.util.concurrent.Executors; import java.text.SimpleDateFormat; class Tasks implements Runnable { private String taskName; // constructor of the class Tasks public Tasks(String str) { // initializing the field taskName taskName = str; } // Printing the task name and then sleeps for 1 sec // The complete process is getting repeated five times public void run() { try { for (int j = 0; j <= 5; j++) { if (j="=" 0) date dt="new" date(); simpledateformat sdf="new" simpledateformat('hh : mm ss'); prints the initialization time for every task system.out.println('initialization name: '+ taskname + '=" + sdf.format(dt)); } else { Date dt = new Date(); SimpleDateFormat sdf = new SimpleDateFormat(" hh execution system.out.println('time of is complete.'); } catch(interruptedexception ie) ie.printstacktrace(); public class threadpoolexample maximum number threads in thread pool static final int max_th="3;" main method void main(string argvs[]) creating five new tasks runnable rb1="new" tasks('task 1'); rb2="new" 2'); rb3="new" 3'); rb4="new" 4'); rb5="new" 5'); a with size fixed executorservice pl="Executors.newFixedThreadPool(MAX_TH);" passes objects to execute (step 3) pl.execute(rb1); pl.execute(rb2); pl.execute(rb3); pl.execute(rb4); pl.execute(rb5); shutdown pl.shutdown(); < pre> <p> <strong>Output:</strong> </p> <pre> Initialization time for the task name: task 1 = 06 : 13 : 02 Initialization time for the task name: task 2 = 06 : 13 : 02 Initialization time for the task name: task 3 = 06 : 13 : 02 Time of execution for the task name: task 1 = 06 : 13 : 04 Time of execution for the task name: task 2 = 06 : 13 : 04 Time of execution for the task name: task 3 = 06 : 13 : 04 Time of execution for the task name: task 1 = 06 : 13 : 05 Time of execution for the task name: task 2 = 06 : 13 : 05 Time of execution for the task name: task 3 = 06 : 13 : 05 Time of execution for the task name: task 1 = 06 : 13 : 06 Time of execution for the task name: task 2 = 06 : 13 : 06 Time of execution for the task name: task 3 = 06 : 13 : 06 Time of execution for the task name: task 1 = 06 : 13 : 07 Time of execution for the task name: task 2 = 06 : 13 : 07 Time of execution for the task name: task 3 = 06 : 13 : 07 Time of execution for the task name: task 1 = 06 : 13 : 08 Time of execution for the task name: task 2 = 06 : 13 : 08 Time of execution for the task name: task 3 = 06 : 13 : 08 task 2 is complete. Initialization time for the task name: task 4 = 06 : 13 : 09 task 1 is complete. Initialization time for the task name: task 5 = 06 : 13 : 09 task 3 is complete. Time of execution for the task name: task 4 = 06 : 13 : 10 Time of execution for the task name: task 5 = 06 : 13 : 10 Time of execution for the task name: task 4 = 06 : 13 : 11 Time of execution for the task name: task 5 = 06 : 13 : 11 Time of execution for the task name: task 4 = 06 : 13 : 12 Time of execution for the task name: task 5 = 06 : 13 : 12 Time of execution for the task name: task 4 = 06 : 13 : 13 Time of execution for the task name: task 5 = 06 : 13 : 13 Time of execution for the task name: task 4 = 06 : 13 : 14 Time of execution for the task name: task 5 = 06 : 13 : 14 task 4 is complete. task 5 is complete. </pre> <p> <strong>Explanation:</strong> It is evident by looking at the output of the program that tasks 4 and 5 are executed only when the thread has an idle thread. Until then, the extra tasks are put in the queue.</p> <p>The takeaway from the above example is when one wants to execute 50 tasks but is not willing to create 50 threads. In such a case, one can create a pool of 10 threads. Thus, 10 out of 50 tasks are assigned, and the rest are put in the queue. Whenever any thread out of 10 threads becomes idle, it picks up the 11<sup>th </sup>task. The other pending tasks are treated the same way.</p> <h2>Risks involved in Thread Pools</h2> <p>The following are the risk involved in the thread pools.</p> <p> <strong>Deadlock:</strong> It is a known fact that deadlock can come in any program that involves multithreading, and a thread pool introduces another scenario of deadlock. Consider a scenario where all the threads that are executing are waiting for the results from the threads that are blocked and waiting in the queue because of the non-availability of threads for the execution.</p> <p> <strong>Thread Leakage:</strong> Leakage of threads occurs when a thread is being removed from the pool to execute a task but is not returning to it after the completion of the task. For example, when a thread throws the exception and the pool class is not able to catch this exception, then the thread exits and reduces the thread pool size by 1. If the same thing repeats a number of times, then there are fair chances that the pool will become empty, and hence, there are no threads available in the pool for executing other requests.</p> <p> <strong>Resource Thrashing:</strong> A lot of time is wasted in context switching among threads when the size of the thread pool is very large. Whenever there are more threads than the optimal number may cause the starvation problem, and it leads to resource thrashing.</p> <h2>Points to Remember</h2> <p>Do not queue the tasks that are concurrently waiting for the results obtained from the other tasks. It may lead to a deadlock situation, as explained above.</p> <p>Care must be taken whenever threads are used for the operation that is long-lived. It may result in the waiting of thread forever and will finally lead to the leakage of the resource.</p> <p>In the end, the thread pool has to be ended explicitly. If it does not happen, then the program continues to execute, and it never ends. Invoke the shutdown() method on the thread pool to terminate the executor. Note that if someone tries to send another task to the executor after shutdown, it will throw a RejectedExecutionException.</p> <p>One needs to understand the tasks to effectively tune the thread pool. If the given tasks are contrasting, then one should look for pools for executing different varieties of tasks so that one can properly tune them.</p> <p>To reduce the probability of running JVM out of memory, one can control the maximum threads that can run in JVM. The thread pool cannot create new threads after it has reached the maximum limit.</p> <p>A thread pool can use the same used thread if the thread has finished its execution. Thus, the time and resources used for the creation of a new thread are saved.</p> <h2>Tuning the Thread Pool</h2> <p>The accurate size of a thread pool is decided by the number of available processors and the type of tasks the threads have to execute. If a system has the P processors that have only got the computation type processes, then the maximum size of the thread pool of P or P + 1 achieves the maximum efficiency. However, the tasks may have to wait for I/O, and in such a scenario, one has to take into consideration the ratio of the waiting time (W) and the service time (S) for the request; resulting in the maximum size of the pool P * (1 + W / S) for the maximum efficiency.</p> <h2>Conclusion</h2> <p>A thread pool is a very handy tool for organizing applications, especially on the server-side. Concept-wise, a thread pool is very easy to comprehend. However, one may have to look at a lot of issues when dealing with a thread pool. It is because the thread pool comes with some risks involved it (risks are discussed above).</p> <hr></=></pre></10;>

stáhněte si tento příklad

Příklad skupiny vláken: 2

Podívejme se na další příklad fondu vláken.

Název souboru: ThreadPoolExample.java

 // important import statements import java.util.Date; import java.util.concurrent.ExecutorService; import java.util.concurrent.Executors; import java.text.SimpleDateFormat; class Tasks implements Runnable { private String taskName; // constructor of the class Tasks public Tasks(String str) { // initializing the field taskName taskName = str; } // Printing the task name and then sleeps for 1 sec // The complete process is getting repeated five times public void run() { try { for (int j = 0; j <= 5; j++) { if (j="=" 0) date dt="new" date(); simpledateformat sdf="new" simpledateformat(\'hh : mm ss\'); prints the initialization time for every task system.out.println(\'initialization name: \'+ taskname + \'=" + sdf.format(dt)); } else { Date dt = new Date(); SimpleDateFormat sdf = new SimpleDateFormat(" hh execution system.out.println(\'time of is complete.\'); } catch(interruptedexception ie) ie.printstacktrace(); public class threadpoolexample maximum number threads in thread pool static final int max_th="3;" main method void main(string argvs[]) creating five new tasks runnable rb1="new" tasks(\'task 1\'); rb2="new" 2\'); rb3="new" 3\'); rb4="new" 4\'); rb5="new" 5\'); a with size fixed executorservice pl="Executors.newFixedThreadPool(MAX_TH);" passes objects to execute (step 3) pl.execute(rb1); pl.execute(rb2); pl.execute(rb3); pl.execute(rb4); pl.execute(rb5); shutdown pl.shutdown(); < pre> <p> <strong>Output:</strong> </p> <pre> Initialization time for the task name: task 1 = 06 : 13 : 02 Initialization time for the task name: task 2 = 06 : 13 : 02 Initialization time for the task name: task 3 = 06 : 13 : 02 Time of execution for the task name: task 1 = 06 : 13 : 04 Time of execution for the task name: task 2 = 06 : 13 : 04 Time of execution for the task name: task 3 = 06 : 13 : 04 Time of execution for the task name: task 1 = 06 : 13 : 05 Time of execution for the task name: task 2 = 06 : 13 : 05 Time of execution for the task name: task 3 = 06 : 13 : 05 Time of execution for the task name: task 1 = 06 : 13 : 06 Time of execution for the task name: task 2 = 06 : 13 : 06 Time of execution for the task name: task 3 = 06 : 13 : 06 Time of execution for the task name: task 1 = 06 : 13 : 07 Time of execution for the task name: task 2 = 06 : 13 : 07 Time of execution for the task name: task 3 = 06 : 13 : 07 Time of execution for the task name: task 1 = 06 : 13 : 08 Time of execution for the task name: task 2 = 06 : 13 : 08 Time of execution for the task name: task 3 = 06 : 13 : 08 task 2 is complete. Initialization time for the task name: task 4 = 06 : 13 : 09 task 1 is complete. Initialization time for the task name: task 5 = 06 : 13 : 09 task 3 is complete. Time of execution for the task name: task 4 = 06 : 13 : 10 Time of execution for the task name: task 5 = 06 : 13 : 10 Time of execution for the task name: task 4 = 06 : 13 : 11 Time of execution for the task name: task 5 = 06 : 13 : 11 Time of execution for the task name: task 4 = 06 : 13 : 12 Time of execution for the task name: task 5 = 06 : 13 : 12 Time of execution for the task name: task 4 = 06 : 13 : 13 Time of execution for the task name: task 5 = 06 : 13 : 13 Time of execution for the task name: task 4 = 06 : 13 : 14 Time of execution for the task name: task 5 = 06 : 13 : 14 task 4 is complete. task 5 is complete. </pre> <p> <strong>Explanation:</strong> It is evident by looking at the output of the program that tasks 4 and 5 are executed only when the thread has an idle thread. Until then, the extra tasks are put in the queue.</p> <p>The takeaway from the above example is when one wants to execute 50 tasks but is not willing to create 50 threads. In such a case, one can create a pool of 10 threads. Thus, 10 out of 50 tasks are assigned, and the rest are put in the queue. Whenever any thread out of 10 threads becomes idle, it picks up the 11<sup>th </sup>task. The other pending tasks are treated the same way.</p> <h2>Risks involved in Thread Pools</h2> <p>The following are the risk involved in the thread pools.</p> <p> <strong>Deadlock:</strong> It is a known fact that deadlock can come in any program that involves multithreading, and a thread pool introduces another scenario of deadlock. Consider a scenario where all the threads that are executing are waiting for the results from the threads that are blocked and waiting in the queue because of the non-availability of threads for the execution.</p> <p> <strong>Thread Leakage:</strong> Leakage of threads occurs when a thread is being removed from the pool to execute a task but is not returning to it after the completion of the task. For example, when a thread throws the exception and the pool class is not able to catch this exception, then the thread exits and reduces the thread pool size by 1. If the same thing repeats a number of times, then there are fair chances that the pool will become empty, and hence, there are no threads available in the pool for executing other requests.</p> <p> <strong>Resource Thrashing:</strong> A lot of time is wasted in context switching among threads when the size of the thread pool is very large. Whenever there are more threads than the optimal number may cause the starvation problem, and it leads to resource thrashing.</p> <h2>Points to Remember</h2> <p>Do not queue the tasks that are concurrently waiting for the results obtained from the other tasks. It may lead to a deadlock situation, as explained above.</p> <p>Care must be taken whenever threads are used for the operation that is long-lived. It may result in the waiting of thread forever and will finally lead to the leakage of the resource.</p> <p>In the end, the thread pool has to be ended explicitly. If it does not happen, then the program continues to execute, and it never ends. Invoke the shutdown() method on the thread pool to terminate the executor. Note that if someone tries to send another task to the executor after shutdown, it will throw a RejectedExecutionException.</p> <p>One needs to understand the tasks to effectively tune the thread pool. If the given tasks are contrasting, then one should look for pools for executing different varieties of tasks so that one can properly tune them.</p> <p>To reduce the probability of running JVM out of memory, one can control the maximum threads that can run in JVM. The thread pool cannot create new threads after it has reached the maximum limit.</p> <p>A thread pool can use the same used thread if the thread has finished its execution. Thus, the time and resources used for the creation of a new thread are saved.</p> <h2>Tuning the Thread Pool</h2> <p>The accurate size of a thread pool is decided by the number of available processors and the type of tasks the threads have to execute. If a system has the P processors that have only got the computation type processes, then the maximum size of the thread pool of P or P + 1 achieves the maximum efficiency. However, the tasks may have to wait for I/O, and in such a scenario, one has to take into consideration the ratio of the waiting time (W) and the service time (S) for the request; resulting in the maximum size of the pool P * (1 + W / S) for the maximum efficiency.</p> <h2>Conclusion</h2> <p>A thread pool is a very handy tool for organizing applications, especially on the server-side. Concept-wise, a thread pool is very easy to comprehend. However, one may have to look at a lot of issues when dealing with a thread pool. It is because the thread pool comes with some risks involved it (risks are discussed above).</p> <hr></=>

Vysvětlení: Při pohledu na výstup programu je zřejmé, že úlohy 4 a 5 se provádějí pouze v případě, že vlákno má nečinné vlákno. Do té doby jsou další úkoly zařazeny do fronty.

Z výše uvedeného příkladu vyplývá, že člověk chce provést 50 úkolů, ale nechce vytvořit 50 vláken. V takovém případě lze vytvořit fond 10 vláken. Takto je přiřazeno 10 z 50 úkolů a zbytek je zařazen do fronty. Kdykoli se kterékoli z 10 vláken stane nečinným, převezme 11^čtúkol. S ostatními nevyřízenými úkoly se zachází stejným způsobem.

designový vzor java

Rizika spojená s fondy vláken

Níže jsou uvedena rizika spojená s fondy vláken.

Zablokování: Je známou skutečností, že uváznutí může nastat v jakémkoli programu, který zahrnuje vícevláknové zpracování, a fond vláken představuje další scénář uváznutí. Zvažte scénář, kde všechna spouštěná vlákna čekají na výsledky z podprocesů, která jsou blokována a čekají ve frontě z důvodu nedostupnosti podprocesů pro spuštění.

Netěsnost závitu: K úniku vláken dochází, když je vlákno odebráno z fondu za účelem provedení úlohy, ale po dokončení úlohy se do něj nevrátí. Když například vlákno vyvolá výjimku a třída fondu není schopna tuto výjimku zachytit, vlákno se ukončí a zmenší velikost fondu vláken o 1. Pokud se totéž opakuje několikrát, pak existuje velká šance, že fond se vyprázdní, a proto ve fondu nejsou k dispozici žádná vlákna pro provádění jiných požadavků.

Thrashing zdrojů: Spousta času se plýtvá přepínáním kontextu mezi vlákny, když je velikost fondu vláken velmi velká. Kdykoli existuje více vláken, než je optimální počet, může to způsobit problém s hladověním a vede to ke ztrátě zdrojů.

Body k zapamatování

Nezařazujte do fronty úlohy, které současně čekají na výsledky získané z jiných úloh. Může to vést k zablokování, jak je vysvětleno výše.

Vždy, když se závity používají pro operaci, která má dlouhou životnost, je třeba dbát opatrnosti. Může to mít za následek věčné čekání vlákna a nakonec povede k úniku zdroje.

Nakonec musí být fond vláken explicitně ukončen. Pokud se tak nestane, program pokračuje v provádění a nikdy nekončí. Vyvoláním metody shutdown() ve fondu vláken ukončíte exekutor. Všimněte si, že pokud se někdo po vypnutí pokusí odeslat exekutorovi další úkol, vyvolá to RejectedExecutionException.

Aby bylo možné efektivně vyladit fond vláken, je třeba porozumět úkolům. Pokud jsou dané úkoly kontrastní, pak by se měl hledat fondy pro provádění různých variant úkolů, aby je bylo možné správně vyladit.

Chcete-li snížit pravděpodobnost, že JVM bude mít nedostatek paměti, je možné řídit maximální počet vláken, která lze v JVM spustit. Fond vláken nemůže vytvářet nová vlákna poté, co dosáhl maximálního limitu.

Fond vláken může používat stejné použité vlákno, pokud vlákno dokončilo své provádění. Tím se ušetří čas a prostředky použité na vytvoření nového vlákna.

Vyladění fondu vláken

O přesné velikosti fondu vláken rozhoduje počet dostupných procesorů a typ úloh, které musí vlákna provádět. Pokud má systém procesory P, které mají pouze procesy typu výpočtu, pak maximální velikost fondu vláken P nebo P + 1 dosahuje maximální účinnosti. Úlohy však mohou čekat na I/O a v takovém scénáři je třeba vzít v úvahu poměr doby čekání (W) a doby služby (S) pro požadavek; výsledkem je maximální velikost bazénu P * (1 + W / S) pro maximální účinnost.

Závěr

Fond vláken je velmi užitečný nástroj pro organizaci aplikací, zejména na straně serveru. Z hlediska koncepce je fond vláken velmi snadno pochopitelný. Při řešení fondu vláken se však možná budete muset podívat na spoustu problémů. Je to proto, že fond vláken s sebou nese určitá rizika (rizika jsou popsána výše).