小编use*_*444的帖子

Process process = Runtime.getRuntime().exec("tasklist");
BufferedReader reader = 
    new BufferedReader(new InputStreamReader(process.getInputStream()));
process.waitFor();

setMaxResults和setFetchSizeorg.hibernate.Query有什么区别？我就是不能得到它=)

我刚刚下载了Eclipse,启动它并发现"Open Project"按钮被禁用.那么如何打开我的项目=)？

当我从jnlp启动应用程序时,我收到消息

" Missing Codebase manifest attribute for:xxx.jar"

这是什么意思？

当我按ctrl- c在控制台中按应用程序线程停止的顺序和关闭挂钩调用？

在Arrays类中,快速排序用于排序基元,但是对于排序对象,它是合并排序.

我想知道为什么会这样？

如果我运行持久的任务,如果第一个任务没有完成,Executor永远不会启动新的线程.有人可以帮我理解为什么以及如何解决这个问题？

import java.util.concurrent.ExecutorService;
import java.util.concurrent.LinkedBlockingQueue;
import java.util.concurrent.ThreadPoolExecutor;
import java.util.concurrent.TimeUnit;

import org.junit.Test;

public class TestExecutor {

    @Test
    public void test() throws InterruptedException {
        ExecutorService checkTasksExecutorService = new ThreadPoolExecutor(1, 10,
                100000, TimeUnit.MILLISECONDS,
                new LinkedBlockingQueue<Runnable>());

        for (int i = 0; i < 20; i++) {
            checkTasksExecutorService.execute(new Runnable() {

                public  void run(){
                    try {
                        System.out.println(Thread.currentThread().getName() + "   running!");
                        Thread.sleep(10000);
                    } catch (Exception e) {
                    }

                }
            });
        }

        Thread.sleep(1000000);
    }
}

在编组期间,我得到了下一个例外

Exception in thread "main" com.sun.xml.internal.ws.encoding.soap.DeserializationException: Failed to read a response: javax.xml.bind.UnmarshalException
 - with linked exception:
[javax.xml.stream.XMLStreamException: ParseError at [row,col]:[1,1127]
Message: XML document structures must start and end within the same entity.]

所以我想看看[行,col]:[1,1127]所在的xml.请建议.

这是生成的hashCode的示例

 @Override
    public int hashCode() {
        int result;
        long temp;
        temp = x != +0.0d ? Double.doubleToLongBits(x) : 0L;
        result = (int) (temp ^ (temp >>> 32));
        temp = y != +0.0d ? Double.doubleToLongBits(y) : 0L;
        result = 31 * result + (int) (temp ^ (temp >>> 32));
        return result;
    }

我想知道那里31*和>>> 32

我有下一个代码

    URL targetUrl = ...
    HttpClient client = new HttpClient(connectionManager);
    GetMethod getMethod = new GetMethod();
    getMethod.setPath(targetUrl.getPath());

    HostConfiguration hostConfiguration = getConfiguration(targetUrl) //unknown lib code

    client.executeMethod(hostConfiguration, getMethod);

在某些情况下(在某些主机上),我得到了

java.lang.IllegalArgumentException: host parameter is null"

在client.executeMethod调用.

为什么会这样？