我有这段代码:
public class BruteForceThread implements Callable<String> {
private static volatile boolean stopped = false;
public String call() {
String password = this.getNextPassword();
while (!stopped) {
System.out.println(numberOfThreat + ": " + password);
synchronized(this) {
try {
if (this.testPassword(password)) {
stopped = true;
return password;
}
} catch (IOException e) {
e.printStackTrace();
}
}
password = this.getNextPassword();
}
return "";
}
}
主要课程:
public static void main(String[] args) throws IOException {
int numberOfThreads = 2;
ExecutorService executor = Executors.newFixedThreadPool(numberOfThreads);
CompletionService<String> pool = new ExecutorCompletionService<String>(executor);
for (int i = 1; i < numberOfThreads + 1; i++) {
Callable<String> bruteForce = new BruteForceThread(...);
pool.submit(bruteForce);
}
executor.shutdown();
for(int i = 0; i < numberOfThreads; i++){
try {
String result = pool.take().get();
if (result != "") {
System.out.println("Your password: " + result);
}
} catch (InterruptedException e) {
e.printStackTrace();
} catch (ExecutionException e) {
e.printStackTrace();
}
}
executor = null;
}
输出: ...
1: aa <- here first thread found a password
2: Fa <- second thread is continue
2: Fb
2: Fc
... many more ...
2: IZ
Your password: aa
如果一个线程找到密码并将stopped设置为true,则另一个线程不会立即停止。为什么呢?
答案 0 :(得分:0)
将有线程等待写入控制台。这意味着当你检查停止和输出到屏幕时,部分或全部线程可能介于两者之间。
BTW除非您正在以高延迟攻击网站,否则您可能会发现破解密码只需要与拥有CPU的线程一样多。更多线程可能会增加开销并降低速度。