Posted on 2007-11-09 01:31
dybjsun 閱讀(213)
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多線程主題
當多線程啟動時，怎么才能控制他們有秩序地執行。本例模擬一個容器，當容器里有東西時，通知各個線程來取得這些東西，如果沒有取到東西，則進入等待狀態。（特別注意在通知各個線程notifyAll和等待wait這些方法一定要寫在同步塊中）
?1?package?com.noahsi.fulltextsearch.index.basic;
?2?
?3?import?java.util.ArrayList;
?4?import?java.util.List;
?5?
?6?import?com.noahsi.fulltextsearch.index.model.LinkModel;
?7?import?com.noahsi.fulltextsearch.util.Debug;
?8?
?9?public?class?Container?{
10?
11??private?List?container?=?null;
12?
13??public?Container()?{
14???container?=?new?ArrayList();
15??}
16?
17??public?Container(LinkModel?model)?{
18???container?=?new?ArrayList();
19???container.add(model);
20??}
21?
22??public?Container(List?list)?{
23???container?=?new?ArrayList();
24???container.addAll(list);
25??}
26?
27??public?synchronized?void?listener()?{
28???if?(container.size()?>?0)?{
29????this.notifyAll();
30???}
31??}
32?
33??public?synchronized?LinkModel?getLinkModel()?{
34???if?(container.size()?==?0)?{
35????try?{
36?????this.wait();
37????}?catch?(InterruptedException?ie)?{
38????}
39????return?null;
40???}
41???return?(LinkModel)?container.remove(0);
42??}
43?
44??public?synchronized?void?putLinksModel(List?links)?{
45???LinkModel?temp?=?null;
46???for?(int?i?=?0;?i?<?links.size();?i++)?{
47????temp?=?(LinkModel)?links.get(i);
48????if?(!container.contains(temp))?{
49?????container.add(temp);
50????}
51???}
52???this.notifyAll();
53??}
54?
55??public?synchronized?void?putLinkModel(LinkModel?model)?{
56???if?(model?!=?null)?{
57????container.add(model);
58???}
59???this.notifyAll();
60??}
61?
62??public?synchronized?void?active()?{
63???this.notifyAll();
64??}
65?
66??public?int?getSize()?{
67???return?container.size();
68??}
69?}