java创建新线程新线程运行,Java中新创建的线程的执行顺序是什么

妖狐艹你老母 2022-10-04 12:53 92阅读 0赞

![Image 1][]

class Test \{

boolean isFirstThread = true;

private synchronized void printer(int threadNo) \{

if(isFirstThread) \{

try \{

Thread.sleep(2000);

\} catch (InterruptedException e) \{

e.printStackTrace();

isFirstThread = false;

System.out.println(threadNo);

public void starter() \{

new Thread()\{

@Override()

public void run() \{

printer(0);

\}.start();

new Thread()\{

@Override()

public void run() \{

printer(1);

\}.start();

new Thread()\{

@Override()

public void run() \{

printer(2);

\}.start();

new Thread()\{

@Override()

public void run() \{

printer(3);

\}.start();

In the above code, when i call starter from main. I have created four new Threads to call a synchronized function. I know the order of execution of the threads can't be predicted. Unless they all wait for some time, so that first thread can finish and come out of the synchronized block. In which case I expect all threads to be held in a queue so i expected the answer as

But consistently(I ran the program more than 20 times) I was getting the output as

Which means that the threads are being held in a stack instead of a queue. Why is it so? Every answer in the google result says it is a queue but I am getting it as a stack. I would like to know the reason behind for holding the threads in stack(which is counter intuitive) instead of queue?

解决方案

The order in which threads start is up to the OS, it is not specified in the Java Language Spec. You call start in the main thread, but when the new thread gets allocated and when it begins processing its Runnable or run method is left to the OS' scheduler to decide.

Be careful not to rely on the order in which threads happen to start.

[Image 1]: