多線程問題,很重要的一點是代碼和數據分離
看了這篇文章,感覺內部類用的很好玩,略有啟發,對其中錯誤的地方修改了下,並去除了取處理器核數和線程池的代碼,重新寫了一個demo。
其實對這個結果並不滿意,因為這個demo中,為各個線程分配任務的方式不太好,太死板,按說應該讓各個線程自己去搶任務。
package learnThread; import java.util.ArrayList; import java.util.List; import java.util.concurrent.Callable; import java.util.concurrent.ExecutionException; import java.util.concurrent.FutureTask; /** * @author lakeslove * 多線程累加求和 * */ public class SumCalculator{ private class Sum implements Callable<Integer>{ private int subMin; private int subMax; public Sum(int subMin,int subMax){ this.subMin = subMin; this.subMax = subMax; } @Override public Integer call() throws Exception { int sum = 0; for(int i = subMin;i <= subMax;i++){ sum += i; System.out.println(Thread.currentThread().getName() + "-------" + sum); } return sum; } } /** * 求和范圍是 min ~ max * @param min * @param max * @param threadNum * @return */ public Integer getSum(int min, int max, int threadNum){ int subMin; int subMax; List<FutureTask<Integer>> taskList = new ArrayList<>(); int sumCounts = max - min + 1; int subCounts = sumCounts/threadNum; int remainder = sumCounts%threadNum; int mark = min; for(int i = 0;i<threadNum;i++){ subMin = mark; if(remainder!=0&&remainder>i){ subMax = subMin + subCounts; }else{ subMax = mark + subCounts - 1; } mark = subMax + 1; System.out.println(subMin +":"+subMax); FutureTask<Integer> task = new FutureTask<Integer>(new Sum(subMin,subMax)); taskList.add(task); new Thread(task).start(); } int sum = taskListSum(taskList); return sum; } private Integer taskListSum(List<FutureTask<Integer>> taskList){ int sum = 0; for(FutureTask<Integer> task : taskList){ try { sum += task.get(); } catch (InterruptedException e) { e.printStackTrace(); } catch (ExecutionException e) { e.printStackTrace(); } } return sum; } /** * @param args * 測試 */ public static void main(String[] args){ SumCalculator sumCalculator = new SumCalculator(); int sum = sumCalculator.getSum(1, 100, 5); System.out.println(sum); } }
