java 多线程
1.题目要求:自定义一个线程,创建三个实例,功能是打印1~100的数字。
其中,数字1+3*n(n>=0)必须是第一个线程实例打印的,2+3*n必须是第二个实例打印,3+3*n必须是第三个线程打印,打印结果严格递增,实例1,2,3依次轮流打印。
2.代码:
public class Printer implements Runnable { int i = 1; public void run() { try { synchronized (this) { while (i <= 100) { String threadNo = Thread.currentThread().getName(); while (i % 3 != Integer.valueOf(threadNo) % 3) { this.wait(); } if (i <= 100) System.out.println(threadNo + ": " + i++); this.notifyAll(); } } } catch (InterruptedException e) { System.out.println("InterruptedException occur"); } } } public class Start { public static void main(String[] args) { Printer printer = new Printer(); Thread t1 = new Thread(printer,"1"); Thread t2 = new Thread(printer,"2"); Thread t3 = new Thread(printer,"3"); t1.start(); t2.start(); t3.start(); } }
代码2:
来个简单粗暴无同步的代码
package cn.byr.nuanyangyang.jishuqi; class PrintingContext { public volatile int current = 1; } class Printer implements Runnable { private PrintingContext ctx; private String name; private int modulo; public Printer(PrintingContext ctx, String name, int modulo) { this.ctx = ctx; this.name = name; this.modulo = modulo; } @Override public void run() { while (true) { int cur = ctx.current; if (cur > 100) { break; } if (cur % 3 == modulo) { System.out.format("[%s] %d\n", name, cur); int newNum = cur + 1; ctx.current = newNum; } } } } public class MangDengDaiJiShuQi { public static void main(String[] args) throws Exception { PrintingContext ctx = new PrintingContext(); Printer p1 = new Printer(ctx, "Printer 1", 1); Printer p2 = new Printer(ctx, "Printer 2", 2); Printer p3 = new Printer(ctx, "Printer 3", 0); Thread t1 = new Thread(p1); Thread t2 = new Thread(p2); Thread t3 = new Thread(p3); long timeStamp1 = System.currentTimeMillis(); t1.start(); t2.start(); t3.start(); t1.join(); t2.join(); t3.join(); long timeStamp2 = System.currentTimeMillis(); System.out.format("Total time: %dms", timeStamp2 - timeStamp1); } }
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