首页 \ 问答 \ 使用ThreadLocal为每个线程分配ID(Assign ID to each thread using ThreadLocal)

使用ThreadLocal为每个线程分配ID(Assign ID to each thread using ThreadLocal)

在下面的程序中,我想为每个线程分配不同的id,但是在输出中,每个线程都有不一致的id,如输出中所示。 但是,如果我取消注释system.out语句,每个线程都被分配了唯一的ID,不知道原因。

class ThreadLocalDemo {
public static void main(String[] args) throws InterruptedException, 
ExecutionException {
    CustomerThread custThread1 = new CustomerThread("Sampath");
    CustomerThread custThread2 = new CustomerThread("Harish");
    CustomerThread custThread3 = new CustomerThread("Harsha");
    CustomerThread custThread4 = new CustomerThread("Gowtham");
    custThread1.start();
    custThread2.start();
    custThread3.start();
    custThread4.start();
    }
}

class CustomerThread extends Thread {
static Integer custId = 0;
private  static ThreadLocal<Integer> tl = new ThreadLocal<Integer>() {
    @Override
    protected Integer initialValue() {
        //System.out.println("will work");
        return ++custId;
    }
};

CustomerThread(String name) {
    super(name);
}

public void run() {
    System.out.println(Thread.currentThread().getName() + " executing with id: " + tl.get());
}
}

输出是:

Sampath executing with id: 1
Harish executing with id: 
Harsha executing with id: 2
Gowtham executing with id: 1

预期的输出是具有唯一ID的线程:

Sampath executing with id: 1
Harish executing with id: 2
Harsha executing with id: 3
Gowtham executing with id: 4              

In the below program i want to assign distinct id to each thread, but in the output each thread is having inconsistent id's as shown in output. However if I uncomment the system.out statement each thread is assigned unique ID's, not sure about the reason.

class ThreadLocalDemo {
public static void main(String[] args) throws InterruptedException, 
ExecutionException {
    CustomerThread custThread1 = new CustomerThread("Sampath");
    CustomerThread custThread2 = new CustomerThread("Harish");
    CustomerThread custThread3 = new CustomerThread("Harsha");
    CustomerThread custThread4 = new CustomerThread("Gowtham");
    custThread1.start();
    custThread2.start();
    custThread3.start();
    custThread4.start();
    }
}

class CustomerThread extends Thread {
static Integer custId = 0;
private  static ThreadLocal<Integer> tl = new ThreadLocal<Integer>() {
    @Override
    protected Integer initialValue() {
        //System.out.println("will work");
        return ++custId;
    }
};

CustomerThread(String name) {
    super(name);
}

public void run() {
    System.out.println(Thread.currentThread().getName() + " executing with id: " + tl.get());
}
}

Output is:

Sampath executing with id: 1
Harish executing with id: 
Harsha executing with id: 2
Gowtham executing with id: 1

Expected output is threads with unique id's:

Sampath executing with id: 1
Harish executing with id: 2
Harsha executing with id: 3
Gowtham executing with id: 4              

原文:https://stackoverflow.com/questions/44338430
更新时间:2023-04-13 13:04

最满意答案

class.getResource(...)将调用classloader.getSystemResource(...)。 它将在用于加载类的相同类路径中查找资源。

您需要将您的gameui.xml添加到您的jar 在清单中的“Class-Path”属性中指定其位置。 这样我们的类加载器就能找到它。

对于选项1 :将资源添加到类路径:为您的jar任务提供一个额外的文件集:

<jar destfile="buildClient/jar/Client.jar" >
<fileset dir="buildClient/classes"/>
<fileset dir="<path to the ui/ dir>" />
...

对于选项2:将目录位置添加到您的类路径中:

<attribute name="Class-Path" value="${lib.list};<path to the ui/ dir>" />

如果你不需要编辑访问这个文件(例如配置你的程序),我会把它放入jar文件中,避免出现外部路径问题。


The class.getResource(...) will call the classloader.getSystemResource(...). It will look for the resource on the same classpath used to load your classes.

You need to add your gameui.xml to your jar or addressed its location in the "Class-Path" attribute in the manifest. That way our classloader will be able to find it.

For option1: Add the resource to the classpath: provide an additional fileset to your jar task:

<jar destfile="buildClient/jar/Client.jar" >
<fileset dir="buildClient/classes"/>
<fileset dir="<path to the ui/ dir>" />
...

For option2: Add the directory location to your classpath:

<attribute name="Class-Path" value="${lib.list};<path to the ui/ dir>" />

If you're don't need edit access to this file (e.g. to configure your program) I would put it into the jar and avoid external path issues.

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