[英]How to return future object from callable interface (internal working of executorService.submit)
[英]ExecutorService.submit(<callable>) taking more time?
我試圖理解java.util.concurrent
包中的實用程序,並了解到我們可以將callable
對象提交給ExecutorService
,它返回Future
,它在call()
成功完成任務后填充了callable
返回的值。方法。
我理解所有的callables都是使用多個線程並發執行的。
當我想看看ExecutorService
對批處理任務執行有多大改進時,我想到了捕獲時間。
以下是我試圖執行的代碼 -
package concurrency;
import java.util.ArrayList;
import java.util.List;
import java.util.concurrent.Callable;
import java.util.concurrent.ExecutionException;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.Future;
public class ExecutorExample {
private static Callable<String> callable = new Callable<String>() {
@Override
public String call() throws Exception {
StringBuilder builder = new StringBuilder();
for(int i=0; i<5; i++) {
builder.append(i);
}
return builder.toString();
}
};
public static void main(String [] args) {
long start = System.currentTimeMillis();
ExecutorService service = Executors.newFixedThreadPool(5);
List<Future<String>> futures = new ArrayList<Future<String>>();
for(int i=0; i<5; i++) {
Future<String> value = service.submit(callable);
futures.add(value);
}
for(Future<String> f : futures) {
try {
System.out.println(f.isDone() + " " + f.get());
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (ExecutionException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
long end = System.currentTimeMillis();
System.out.println("Executer callable time - " + (end - start));
service.shutdown();
start = System.currentTimeMillis();
for(int i=0; i<5; i++) {
StringBuilder builder = new StringBuilder();
for(int j=0; j<5; j++) {
builder.append(j);
}
System.out.println(builder.toString());
}
end = System.currentTimeMillis();
System.out.println("Normal time - " + (end - start));
}
}
這是這個的輸出 -
true 01234
true 01234
true 01234
true 01234
true 01234
Executer callable time - 5
01234
01234
01234
01234
01234
Normal time - 0
如果我遺漏了某些東西或以錯誤的方式理解某些東西,請告訴我。
在此先感謝您的時間和幫助。
如果您在Callable中的任務很小,那么您將無法從並發到期任務切換和初始化開銷中獲益。 嘗試在可調用中添加更多更重的循環,比如1000000次迭代,你可以看到差異
當您第一次運行任何代碼esp時,需要時間。 如果將任務傳遞給另一個線程,則可能需要1-10微秒,如果您的任務花費的時間少於此,則開銷可能大於優勢。 即如果您的開銷足夠高,使用多個線程可能比使用單個線程慢得多。
我建議你
不是答案(但我不確定代碼是否符合評論)。 為了擴展彼得所說的內容,通常可以找到工作規模(以執行時間衡量)的最佳位置,以平衡工作人員之間的公平工作分配的池/隊列開銷。 代碼示例有助於找到該最佳位置的估計值。 在目標硬件上運行。
import java.util.concurrent.*;
import java.util.concurrent.atomic.*;
public class FibonacciFork extends RecursiveTask<Long> {
private static final long serialVersionUID = 1L;
public FibonacciFork( long n) {
super();
this.n = n;
}
static ForkJoinPool fjp = new ForkJoinPool( Runtime.getRuntime().availableProcessors());
static long fibonacci0( long n) {
if ( n < 2) {
return n;
}
return fibonacci0( n - 1) + fibonacci0( n - 2);
}
static int rekLimit = 8;
private static long stealCount;
long n;
private long forkCount;
private static AtomicLong forks = new AtomicLong( 0);
public static void main( String[] args) {
int n = 45;
long times[] = getSingleThreadNanos( n);
System.out.println( "Single Thread Times complete");
for ( int r = 2; r <= n; r++) {
runWithRecursionLimit( r, n, times[ r]);
}
}
private static long[] getSingleThreadNanos( int n) {
final long times[] = new long[ n + 1];
ExecutorService es = Executors.newFixedThreadPool( Math.max( 1, Runtime.getRuntime().availableProcessors() / 2));
for ( int i = 2; i <= n; i++) {
final int arg = i;
Runnable runner = new Runnable() {
@Override
public void run() {
long start = System.nanoTime();
final int minRuntime = 1000000000;
long runUntil = start + minRuntime;
long result = fibonacci0( arg);
long end = System.nanoTime();
int ntimes = Math.max( 1, ( int) ( minRuntime / ( end - start)));
if ( ntimes > 1) {
start = System.nanoTime();
for ( int i = 0; i < ntimes; i++) {
result = fibonacci0( arg);
}
end = System.nanoTime();
}
times[ arg] = ( end - start) / ntimes;
}
};
es.execute( runner);
}
es.shutdown();
try {
es.awaitTermination( 1, TimeUnit.HOURS);
} catch ( InterruptedException e) {
System.out.println( "Single Timeout");
}
return times;
}
private static void runWithRecursionLimit( int r, int arg, long singleThreadNanos) {
rekLimit = r;
long start = System.currentTimeMillis();
long result = fibonacci( arg);
long end = System.currentTimeMillis();
// Steals zählen
long currentSteals = fjp.getStealCount();
long newSteals = currentSteals - stealCount;
stealCount = currentSteals;
long forksCount = forks.getAndSet( 0);
System.out.println( "Fib(" + arg + ")=" + result + " in " + ( end-start) + "ms, recursion limit: " + r +
" at " + ( singleThreadNanos / 1e6) + "ms, steals: " + newSteals + " forks " + forksCount);
}
static long fibonacci( final long arg) {
FibonacciFork task = new FibonacciFork( arg);
long result = fjp.invoke( task);
forks.set( task.forkCount);
return result;
}
@Override
protected Long compute() {
if ( n <= rekLimit) {
return fibonacci0( n);
}
FibonacciFork ff1 = new FibonacciFork( n-1);
FibonacciFork ff2 = new FibonacciFork( n-2);
ff1.fork();
long r2 = ff2.compute();
long r1 = ff1.join();
forkCount = ff2.forkCount + ff1.forkCount + 1;
return r1 + r2;
}
}
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