我想用 T1-1、T2-2、T3-3、T1-4、T2-5、T3-6 等線程按順序打印數(shù)字public class NumberGame { static int a=1; public static void main(String args[]) throws InterruptedException { PrintSequenceRunnable C1=new PrintSequenceRunnable("T1",a); PrintSequenceRunnable C2=new PrintSequenceRunnable("T2",a); PrintSequenceRunnable C3=new PrintSequenceRunnable("T3",a); Thread t1 = new Thread(C1); Thread t2 = new Thread(C2); Thread t3 = new Thread(C3); t1.start(); t2.start(); t3.start(); }}public class PrintSequenceRunnable implements Runnable { String tname; int a; PrintSequenceRunnable(String tname, int a ) { this.tname = tname; this.a = a; } @Override public void run() { synchronized (this) { for(int i=0; i<10;i++) { System.out.println(tname+" "+a); a++; try { this.wait(1000); this.notify(); } catch (InterruptedException e) { // TODO Auto-generated catch block e.printStackTrace(); } } } // TODO Auto-generated method stub }}但我的輸出就像T1-1 T2-1 T3-1 T1-2 T3-2 T2-2 T3-3 T1-3 T2-3 T3-4 T1-4 T2-4 T3-5 T1-5 T2-5 T3-6 T1- 6 T2-6 T1-7 T2-7 T3-7 T2-8 T3-8 T1-8 T2-9 T3-9 T1-9 T2-10 T1-10 T3-10誰(shuí)能幫我。
2 回答
慕田峪7331174
TA貢獻(xiàn)1828條經(jīng)驗(yàn) 獲得超13個(gè)贊
問題是：
該設(shè)計(jì)存在線程之間引發(fā)條件的問題，您需要同步它們。
在使用構(gòu)造函數(shù)時(shí)，
PrintSequenceRunnable(String tname, int a )您將發(fā)送原始變量的副本a，它是靜態(tài)成員NumberGame。所以，每個(gè)PrintSequenceRunnable都有自己的變量a。
我的建議是使用方法wait和同步每個(gè)線程notify。我拿了你的代碼并做了一些修改：
數(shù)字游戲
public class NumberGame {
public static void main(String args[]) throws InterruptedException
{
PrintSequenceRunnable C1=new PrintSequenceRunnable("T1");
PrintSequenceRunnable C2=new PrintSequenceRunnable("T2");
PrintSequenceRunnable C3=new PrintSequenceRunnable("T3");
Thread t1 = new Thread(C1);
Thread t2 = new Thread(C2);
Thread t3 = new Thread(C3);
t1.start();
t2.start();
t3.start();
Thread.sleep(1);//Wait 1 ms to avoid a raise condition
PrintSequenceRunnable.activateNextItem(); //Start sequence.
t1.join();
t2.join();
t3.join();
System.out.println("--END--");
}
}
打印序列可運(yùn)行
import java.util.Vector;
public class PrintSequenceRunnable implements Runnable {
static private int a = 0;
private static Vector<PrintSequenceRunnable> items = new Vector<PrintSequenceRunnable>();
/**
* Method to select the next Thread which will be activate to continue its thread.
*/
public static synchronized void activateNextItem() {
int index = a % items.size();
items.get(index).activate();
}
private String tname;
private Object sempahoro = new Object(); //Object to sinchrony the thread
public PrintSequenceRunnable(String tname)
{
this.tname = tname;
items.add(this);
}
public void activate()
{
synchronized (sempahoro) {
sempahoro.notify();
}
}
@Override
public void run() {
for(int i=0; i<10;i++)
{
synchronized (sempahoro) {
try {
sempahoro.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
a++;
System.out.println(tname+" "+a);
activateNextItem(); //Raise the next thread.
}
// TODO Auto-generated method stub
}
}
在此示例中，方法activateNextItem, fromPrintSequenceRunnable` 將決定通知哪個(gè)實(shí)例執(zhí)行其線程。
重要的是，我需要sleep在初始化每個(gè)線程后設(shè)置一秒以避免引發(fā)條件，我的意思是：等待所有線程啟動(dòng)并使所有線程處于等待狀態(tài)。
輸出：
T1 1
T2 2
T3 3
T1 4
T2 5
T3 6
T1 7
T2 8
T3 9
T1 10
T2 11
T3 12
T1 13
T2 14
T3 15
T1 16
T2 17
T3 18
T1 19
T2 20
T3 21
T1 22
T2 23
T3 24
T1 25
T2 26
T3 27
T1 28
T2 29
T3 30
--END--
慕少森
TA貢獻(xiàn)2019條經(jīng)驗(yàn) 獲得超9個(gè)贊
只是為了學(xué)習(xí)，這就是強(qiáng)制順序輸出的方法。但請(qǐng)注意，不能以任何方式保證線程的順序執(zhí)行。正如其他人指出的那樣，這個(gè)問題不適合多線程處理。如果您想按順序執(zhí)行某件事，請(qǐng)?jiān)谝粋€(gè)線程中執(zhí)行。
public class NumberGame {
public static void main(String[] args) {
PrintSequenceRunnable.startFrom("T1");
new Thread(new PrintSequenceRunnable("T1", "T2")).start();
new Thread(new PrintSequenceRunnable("T2", "T3")).start();
new Thread(new PrintSequenceRunnable("T3", "T1")).start();
}
}
class PrintSequenceRunnable implements Runnable {
private final String name;
private final String next;
private static String moveTo;
private static int value = 1;
PrintSequenceRunnable(String name, String next) {
this.name = name;
this.next = next;
}
static void startFrom(String start) {
moveTo = start;
}
private int uselessCounter = 0;
@Override
public void run() {
do {
synchronized (moveTo) {
if (name.equals(moveTo)) {
System.out.println(name + "-" + (value++));
moveTo = next;
} else {
uselessCounter++;
}
}
} while (value < 10);
System.out.println("Ran " + name + " uselessly for " + uselessCounter + " times."); // remove it.
}
}
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