如何进行jdk7新特性中fork/join框架的原理分析,相信很多没有经验的人对此束手无策,为此本文总结了问题出现的原因和解决方法,通过这篇文章希望你能解决这个问题。原理解析:fork分解,join结合。这个框架的本质是将一个任务分解成多个
如何进行jdk7新特性中fork/join框架的原理分析,相信很多没有经验的人对此束手无策,为此本文总结了问题出现的原因和解决方法,通过这篇文章希望你能解决这个问题。
原理解析:fork分解,join结合。这个框架的本质是将一个任务分解成多个子任务,每个子任务用单独的线程去处理。这里用到了递归的思想。框架的结构图可以参考
使用fork/join 框架很简单,
实现子问题的一般求解算法
如何分解问题
继承 RecursiveAction ,实现compute()方法
伪代码代码
Result solve(Problem problem) { if (problem is small) directly solve problem else { split problem into independent parts fork new subtasks to solve each part join all subtasks compose result from subresults }
这里我通过一个改进的二分查找来讲解fork/join的使用。(后面才发现,选用这个案例是非常失败的,因为二分查找的时间是logn,而创建线程的开销更大,这样并不能体现多线程二分查找的优势,所以这个代码不具有实用性,只是为了说明如何使用框架:)
代码如下:
BinarySearchProblem.java
Java代码
package testjdk7; import java.util.Arrays; public class BinarySearchProblem { private final int[] numbers; private final int start; private final int end; public final int size; public BinarySearchProblem(int[] numbers,int start,int end){ this.numbers = numbers; this.start = start; this.end = end; this.size = end -start; } public int searchSequentially(int numberToSearch){ //偷懒,不自己写二分查找了 return Arrays.binarySearch(numbers, start, end, numberToSearch); } public BinarySearchProblem subProblem(int subStart,int subEnd){ return new BinarySearchProblem(numbers,start+subStart,start+subEnd); } }
BiSearchWithForkJoin.java
Java代码
package testjdk7; import java.util.concurrent.ForkJoinPool; import java.util.concurrent.RecursiveAction; public class BiSearchWithForkJoin extends RecursiveAction { private final int threshold; private final BinarySearchProblem problem; public int result; private final int numberToSearch; public BiSearchWithForkJoin(BinarySearchProblem problem,int threshold,int numberToSearch){ this.problem = problem; this.threshold = threshold; this.numberToSearch = numberToSearch; } @Override protected void compute() { if(problem.size < threshold){ //小于阀值,就直接用普通的二分查找 result = problem.searchSequentially(numberToSearch); }else{ //分解子任务 int midPoint = problem.size/2; BiSearchWithForkJoin left = new BiSearchWithForkJoin(problem.subProblem(0, midPoint),threshold,numberToSearch); BiSearchWithForkJoin right = new BiSearchWithForkJoin(problem.subProblem(midPoint+1, problem.size),threshold,numberToSearch); invokeAll(left,right); result = Math.max(left.result, right.result); } } //构造数据 private static final int[] data = new int[1000_0000]; static{ for(int i = 0;i<1000_0000;i++){ data[i] = i; } } public static void main(String[] args){ BinarySearchProblem problem = new BinarySearchProblem(data,0,data.length); int threshold = 100; int nThreads = 10; //查找100_0000所在的下标 BiSearchWithForkJoin bswfj = new BiSearchWithForkJoin(problem,threshold,100_0000); ForkJoinPool fjPool = new ForkJoinPool(nThreads); fjPool.invoke(bswfj); System.out.printf("Result is:%d%n",bswfj.result); } }
RecursiveTask 还可以带返回值,这里给出一段代码作为参考(斐波那契函数)
(来自Http://www.ibm.com/developerworks/cn/java/j-lo-forkjoin/index.html)
Java代码
class Fibonacci extends RecursiveTask { final int n; Fibonacci(int n) { this.n = n; } private int compute(int small) { final int[] results = { 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89 }; return results[small]; } public Integer compute() { if (n <= 10) { return compute(n); } Fibonacci f1 = new Fibonacci(n - 1); Fibonacci f2 = new Fibonacci(n - 2); System.out.println("fork new thread for " + (n - 1)); f1.fork(); System.out.println("fork new thread for " + (n - 2)); f2.fork(); return f1.join() + f2.join(); } }
用途
只要问题能够分解成类似子问题的,都可以使用这个框架。对于大批量的数据尤其合适。
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