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[英]Is the below is correct equals and hashCode implementation in Java?
[英]is the implementation of the Recursive Task below correct?
我開始了解遞歸任務和遞歸操作的實現。 根據我的理解和一些 java 文檔,我想出了以下代碼來將數組中的所有數字相加。
我需要幫助來糾正這個問題,並幫助我指出我哪里出錯了。
import java.util.concurrent.ForkJoinPool;
import java.util.concurrent.ForkJoinTask;
import java.util.concurrent.RecursiveTask;
public class ForkJoinPoolTest {
public static void main(String[] args) {
ForkJoinPool pool = new ForkJoinPool(4);
long[] numbers = {1,2,3,4,5,6,7,8,9};
AdditionTask newTask = new AdditionTask(numbers, 0, numbers.length -1 );
ForkJoinTask<Long> submit = pool.submit(newTask);
System.out.println(submit.join());
}
}
class AdditionTask extends RecursiveTask<Long> {
long[] numbers;
int start;
int end;
public AdditionTask(long[] numbers, int start, int end) {
this.numbers = numbers;
this.start = start;
this.end = end;
}
@Override
protected Long compute() {
if ((end - start) > 2) {
int length = numbers.length;
int mid = (length % 2 == 0) ? length / 2 : (length - 1) / 2;
AdditionTask leftSide = new AdditionTask(numbers, 0, mid);
leftSide.fork();
AdditionTask rightSide = new AdditionTask(numbers, mid+1, length-1);
return rightSide.compute() + leftSide.join();
} else {
return numbers[0] + numbers[1];
}
}
}
新代碼 [已修復]這是我修復的代碼,似乎僅適用於小數組。 在下面的示例中,數組大小為 10000,總和是錯誤的。 為什么它計算錯誤的總和?
import java.util.Random;
import java.util.concurrent.ForkJoinPool;
import java.util.concurrent.ForkJoinTask;
import java.util.concurrent.RecursiveTask;
public class ForkJoinPoolTest {
public static void main(String[] args) {
Random r = new Random();
int low = 10000;
int high = 100000;
int size = 100000;
long[] numbers = new long[size];
int sum = 0;
for (int i = 0; i < size; i++) {
int n = r.nextInt(high - low) + low;
numbers[i] = n;
sum += numbers[i];
}
long s = System.currentTimeMillis();
ForkJoinPool pool = new ForkJoinPool(1);
AdditionTask newTask = new AdditionTask(numbers, 0, numbers.length-1);
ForkJoinTask<Long> submit = pool.submit(newTask);
System.out.println("Expected Answer: " + sum + ", Actual: " + submit.join());
long e = System.currentTimeMillis();
System.out.println("Total time taken: " + (e - s) + " ms in parallel Operation");
long s2 = System.currentTimeMillis();
System.out.println("Started: " + s2);
int manualSum = 0;
for (long number : numbers) {
manualSum += number;
}
System.out.println("Expected Answer: " + sum + ", Actual: " + manualSum);
long e2 = System.currentTimeMillis();
System.out.println("Ended: " + e2);
System.out.println("Total time taken: " + (e2 - s2) + " ms in sequential Operation");
}
}
class AdditionTask extends RecursiveTask<Long> {
long[] numbers;
int start;
int end;
public AdditionTask(long[] numbers, int start, int end) {
this.numbers = numbers;
this.start = start;
this.end = end;
}
@Override
protected Long compute() {
int length = (start == 0) ? end +1 : (end - (start - 1));
if (length > 2) {
int mid = (length % 2 == 0) ? length / 2 : (length - 1) / 2;
AdditionTask leftSide = new AdditionTask(numbers, start, (start+mid));
leftSide.fork();
AdditionTask rightSide = new AdditionTask(numbers, (start+mid)+1, end);
Long rightSideLong = rightSide.compute();
Long leftSideLong = leftSide.join();
Long total = rightSideLong + leftSideLong;
return total;
} else {
if (start == end) {
return numbers[start];
}
return numbers[start] + numbers[end];
}
}
}
您的並行計算的第二個版本是正確的。 但是代碼中的兩個非並行計算都被破壞了,因為它們使用int
作為它們的總和,這對於大型數組會溢出。 當您修復它們時,也使用long
,它們將產生與您的並行計算相同的結果。
盡管如此,還是有一些需要改進的地方。 首先,你應該擺脫這些條件:
int length = (start == 0) ? end +1 : (end - (start - 1));
和
int mid = (length % 2 == 0) ? length / 2 : (length - 1) / 2;
它們沒有提供比更簡單的好處
int length = end - (start - 1); // or end - start + 1
和
int mid = length / 2;
那么,一個高效的並行處理不應該盡可能多地分解,而是結合實際可實現的並行性。 您可以getSurplusQueuedTaskCount()
使用getSurplusQueuedTaskCount()
@Override
protected Long compute() {
int length = end - (start - 1);
// only split when benefit from parallel processing is likely
if (length > 2 && getSurplusQueuedTaskCount() < 2) {
int mid = length / 2;
AdditionTask leftSide = new AdditionTask(numbers, start, (start+mid));
leftSide.fork();
AdditionTask rightSide = new AdditionTask(numbers, (start+mid)+1, end);
Long rightSideLong = rightSide.compute();
// do in this thread if no worker thread has picked it up yet
Long leftSideLong = leftSide.tryUnfork()? leftSide.compute(): leftSide.join();
Long total = rightSideLong + leftSideLong;
return total;
} else { // do sequential
long sum = 0;
for(int ix = start; ix <= end; ix++) sum += numbers[ix];
return sum;
}
}
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