我正在寻找在线程系统（主从）中使用数组的解决方案，它允许我通过用户输入在多个线程上划分矩阵的计算，并通过1个主线程将其指向计算1的多个从属线程矩阵的领域。
我试图运用我的知识，但我只是在方法中转移问题。 代码有效，但只使用1个线程（接一个）而不是同时使用。
以为我可以用某种方式划分矩阵（这是一个Spliterator吗？）但不知道究竟是怎么回事。
如果结果字段为空，也尝试使用布尔值但是它根本不起作用。
import java.io.BufferedReader;
import java.io.File;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Scanner;
class Threadverteiler extends Thread {
public Threadverteiler(Thread[] threads) {
synchronisiert(threads);
}
public void synchronisiert(Thread[] t) {
Thread[] threads = t;
synchronized (threads) {
for (int i = 0; i < threads.length; i++) {
threads[i].start();
System.out.println("Thread " + i + " gestartet");
} // TODO Auto-generated constructor stub
}
}
}
class Threads extends Thread {
public Threads(int[][] a, int[][] b) {
run(a, b);
}
/*
* public Threadverteiler(Thread[] c, int[][] a,int[][]b) { //start();
* run(a, b); //Arrays.spliterator(a);
* //System.out.println(Arrays.spliterator(a));
*
* }
*/
public void run(int[][] a, int[][] b) {
// synchronized (this) {
// boolean arrayleer = false;
// int[][]cc= new int[5][5];
// public int[][] rechnen(int[][] a,int[][]b){
// while( arrayleer==true){
int[][] aa = new int[5][5];
int[][] bb = new int[5][5];
int[][] cc = new int[5][5];
aa = a;
bb = b;
for (int i = 0; i < aa.length; i++)
{
for (int j = 0; j < bb.length; j++)
{
for (int k = 0; k < cc.length; k++)
{
cc[i][j] = cc[i][j] + aa[i][k] * bb[k][j];
// cc[i][j] = cc[i][j] + a[i][k] * b[k][j];
/*
* if(cc[i][j]==0){ i++; j++; k++; continue; } else {
* System.out.println("Andere thread"); break;
* //arrayleer=false; }
*/
}
}
}
/*
* try { PrintWriter print = new PrintWriter(new File());
*
* } catch (Exception e) { // TODO: handle exception } return cc;
*
*
* } }
*/
// c=cc;
// System.out.println("active Threads " + activeCount());
System.out.println(Arrays.deepToString(cc));
System.out.println("active Threads " + activeCount());
}
}
// }
public class Uebung2 {
public static void main(String[] args) {
int[][] a = { { 1, -2, 3, 4, -1 }, { -2, 3, 0, 1, 2 }, { 4, -1, 2, 1, -2 }, { -2, 1, 3, -1, 3 },
{ 0, 2, -1, 2, 4 } };
int[][] b = { { 2, -4, -1, 1, -2 }, { -1, 1, -2, 2, 1 }, { 5, 0, 3, -2, -4 }, { 1, -2, 1, 0, 2 },
{ 2, 3, -3, 0, 0 } };
// int[][] c= new int[5][5];
// System.out.println(Arrays.deepToString(a));
// System.out.println(Arrays.deepToString(b));
// public static void threadauswahl(int x){
// int i = 0;
System.out.println("Bitte Anzahl Threads eingeben");
Scanner sc = new Scanner(System.in);
int eingabe = sc.nextInt();
sc.close();
Thread[] threads = new Thread[eingabe];
// while (threads[i]==null) {
for (int i = 0; i < threads.length; i++) {
threads[i] = new Thread(new Threads(a, b));
// threads[i].start(); // !!!!!!!!!!!!!!!!! ändern!!
// i++;
}
Threadverteiler s = new Threadverteiler(threads);
}
}
// System.out.println(Arrays.deepToString(a));
// System.out.println(Arrays.deepToString(b));
// System.out.println(Arrays.deepToString(c));
答案 0 :(得分：0)
如果我正确理解你的问题，下面应该启动多个线程
public class Main {
public static void main(final String[] args) {
final int[][] a = { { 1, -2, 3, 4, -1 }, { -2, 3, 0, 1, 2 }, { 4, -1, 2, 1, -2 }, { -2, 1, 3, -1, 3 }, { 0, 2, -1, 2, 4 } };
final int[][] b = { { 2, -4, -1, 1, -2 }, { -1, 1, -2, 2, 1 }, { 5, 0, 3, -2, -4 }, { 1, -2, 1, 0, 2 }, { 2, 3, -3, 0, 0 } };
System.out.println("Bitte Anzahl Threads eingeben");
final Scanner sc = new Scanner(System.in);
final int eingabe = sc.nextInt();
sc.close();
final Thread[] threads = new Thread[eingabe];
for (int i = 0; i < threads.length; i++) {
threads[i] = new Thread(new Threads(a, b));
}
final Threadverteiler s = new Threadverteiler(threads);
}
}
class Threadverteiler extends Thread {
public Threadverteiler(final Thread[] threads) {
synchronisiert(threads);
}
public void synchronisiert(final Thread[] t) {
final Thread[] threads = t;
synchronized (threads) {
for (int i = 0; i < threads.length; i++) {
threads[i].start();
System.out.println("Thread " + i + " gestartet");
}
for (int i = 0; i < threads.length; i++) {
try {
threads[i].join();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
}
class Threads extends Thread {
private final int[][] a;
private final int[][] b;
public Threads(final int[][] a, final int[][] b) {
this.a = a;
this.b = b;
}
@Override
public void run() {
int[][] aa = new int[5][5];
int[][] bb = new int[5][5];
final int[][] cc = new int[5][5];
aa = a;
bb = b;
for (int i = 0; i < aa.length; i++) {
for (int j = 0; j < bb.length; j++) {
for (int k = 0; k < cc.length; k++) {
cc[i][j] = cc[i][j] + aa[i][k] * bb[k][j];
}
}
}
System.out.println(Arrays.deepToString(cc));
System.out.println("active Threads " + activeCount());
}
}
答案 1 :(得分：0)
如果我理解正确，此设计已在API中通过ExecutorService和Future实施。这些将允许您（ExecutorService）控制一个任务列表，每个任务包含一个矩阵段，而Futures会在完成时报告。你仍然需要为每个Future任务分解矩阵，但这应该是微不足道的。