JDK5.0 ThreadPool 多线程计算

package karl.test.threadpool;

import java.util.concurrent.Callable;
import java.util.concurrent.CompletionService;
import java.util.concurrent.ExecutionException;
import java.util.concurrent.ExecutorCompletionService;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;

public  class ConcurrentCalculator2 {
     private ExecutorService exec;

     private CompletionService<Long> completionService;

     private  int cpuCoreNumber;

     class SumCalculator  implements Callable<Long> {
         private  int[] numbers;

         private  int start;

         private  int end;

         public SumCalculator( int[] numbers,  int start,  int end) {
             this.numbers = numbers;
             this.start = start;
             this.end = end;
        }

         public Long call()  throws Exception {
            Long sum = 0L;
             for ( int i = start; i < end; i++) {
                sum += numbers[i];
            }
             return sum;
        }

    }

     public ConcurrentCalculator2() {
        cpuCoreNumber = Runtime.getRuntime().availableProcessors();
        exec = Executors.newFixedThreadPool(cpuCoreNumber);
        completionService =  new ExecutorCompletionService<Long>(exec);
    }

     public Long sum( final  int[] numbers) {
         for ( int i = 0; i < cpuCoreNumber; i++) {
             int increment = numbers.length / cpuCoreNumber + 1;
             int start = increment * i;
             int end = increment * i + increment;
             if (end > numbers.length) {
                end = numbers.length;
            }
            SumCalculator subCalc =  new SumCalculator(numbers, start, end);
             if (!exec.isShutdown()) {
                completionService.submit(subCalc);
            }
        }
         return getResult();
    }

     public Long getResult() {
        Long result = 0L;
         for ( int i = 0; i < cpuCoreNumber; i++) {
             try {
                Long subSum = completionService.take().get();//take()已经完成的任务，如果暂时还没有完成的任务，take()方法也会阻塞。
                result += subSum;
            }  catch (InterruptedException e) {
            }  catch (ExecutionException e) {
            }
        }
         return result;
    }

     public  void close() {
        exec.shutdown();
    }

     public  static  void main(String[] args) {
         int[] numbers =  new  int[] { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 };
        ConcurrentCalculator2 calc =  new ConcurrentCalculator2();
        Long sum = calc.sum(numbers);
        System.out.println(sum);
        calc.close();
    }
}