What's the best way to quit a Java application with code?
You can use System.exit() for this purpose.
According to oracle's Java 8 documentation:
public static void exit(int status)
Terminates the currently running Java Virtual Machine. The argument serves as a status code; by convention, a nonzero status code indicates abnormal termination.
This method calls the exit method in class Runtime. This method never returns normally.
The call System.exit(n) is effectively equivalent to the call:
Runtime.getRuntime().exit(n)
System.exit(0);
The "0" lets whomever called your program know that everything went OK. If, however, you are quitting due to an error, you should System.exit(1);, or with another non-zero number corresponding to the specific error.
Also, as others have mentioned, clean up first! That involves closing files and other open resources.
System.exit(int i) is to be used, but I would include it inside a more generic shutdown() method, where you would include "cleanup" steps as well, closing socket connections, file descriptors, then exiting with System.exit(x).
System.exit() is usually not the best way, but it depends on your application.
The usual way of ending an application is by exiting the main() method. This does not work when there are other non-deamon threads running, as is usual for applications with a graphical user interface (AWT, Swing etc.). For these applications, you either find a way to end the GUI event loop (don't know if that is possible with the AWT or Swing), or invoke System.exit().
Using dispose(); is a very effective way for closing your programs.
I found that using System.exit(x) resets the interactions pane and supposing you need some of the information there it all disappears.
I agree with Jon, have your application react to something and call System.exit().
Be sure that:
you use the appropriate exit value. 0 is normal exit, anything else indicates there was an error
you close all input and output streams. Files, network connections, etc.
you log or print a reason for exiting especially if its because of an error
The answer is System.exit(), but not a good thing to do as this aborts the program. Any cleaning up, destroy that you intend to do will not happen.
There's two simple answers to the question.
This is the "Professional way":
//This just terminates the program.
System.exit(0);
This is a more clumsier way:
//This just terminates the program, just like System.exit(0).
return;
Runtime.getCurrentRumtime().halt(0);
System.exit() will do what you want. But in most situations, you probably want to exit a thread, and leave the main thread alive. By doing that, you can terminate a task, but also keep the ability to start another task without restarting the app.
System.exit(ABORT);
Quit's the process immediately.
This should do it in the correct way:
mainFrame.setDefaultCloseOperation(JFrame.DO_NOTHING_ON_CLOSE);
mainFrame.addWindowListener(new WindowListener() {
#Override
public void windowClosing(WindowEvent e) {
if (doQuestion("Really want to exit?")) {
mainFrame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
mainFrame.dispose();
}
}
Related
How do I make java quit the program when it is triggered? I have tried pretty much everything I can do, but I just can't make it quit.
You can use System.exit(0); where 0 is the exit code. This works on all platforms (including android).
Here is a link to the official documentation if you require further reading:
Oracle Documentation
public static void exit(int status)
Terminates the currently running Java Virtual Machine. The argument serves as a status code; by convention, a nonzero status code indicates abnormal termination. This method calls the exit method in class Runtime. This method never returns normally.
The call System.exit(n) is effectively equivalent to the call: Runtime.getRuntime().exit(n)
I'm looking for a way to break into the debugger from Java code, without setting a breakpoint in the IDE.
In Win32 there was DebugBreak(), in C# there's DebugBreak() Equivalent in C#, but I can't find anything in Java.
What I'm trying to do: say I have a wait with a 30s timeout, but in normal conditions that wait should always be <1s. I'd like to use ByteMan or something similar to wait with 1s timeout first, and break into the debugger if that wait timed out.
Not a direct answer to your question but in most IDE's you can set conditional breakpoints.
This would require you to set a timestamp variable in your code before the timeout and test its value in your breakpoint, after the timeout. You would trigger your breakpoint conditional if the delta is greater than your threshold.
The poster of this question has done exactly what I was looking for: Secure Debugging for Production JVMs
It reproduces the best feature of DebugBreak, that is you can attach the IDE to your program only after the "breakpoint" is hit.
I created this class
import com.sun.jna.Library;
public interface Kernel32 extends Library {
public void OutputDebugStringA(String Text);
public void DebugBreak();
}
then used it like this:
Kernel32 lib = (Kernel32) Native.loadLibrary("kernel32", Kernel32.class);
lib.OutputDebugStringA("just before DebugBreak\n");
lib.DebugBreak();
and that's it. It calls native function
void WINAPI DebugBreak(void);
inside Kernel32.dll
You also need the jar file jna-4.4.0.jar
for easier situation inside IDE use also this example with breakpoint in body, poor mans conditional... (sorry for c# format)
if (lineNumber==6502)
{
System.out.println("BREAK"); //breakpoint here
}
Currently (as of 2021) you cannot accomplish this in Java in a way as convenient as DebugBreak is in C#.
The best you can do in Java is throw a custom runtime exception (say, class DebugBreakException extends RuntimeException) and pray that you have also remembered to add an "exception breakpoint" which breaks whenever this exception is thrown, even if the exception is handled.
Immediately after your code throws this exception, it catches it and dumps a message saying that this particular exception occurred.
If this message ever appears in the log, and yet your debugger did not break, then you know you forgot to add the necessary "exception breakpoint." (In other words, you have to pray that if this message ever gets logged, it will be noticed by someone.)
You can also be extra smart about it and add some code which checks how many milliseconds have elapsed from the moment the exception was thrown until it was caught; if more than, say, 10 milliseconds elapsed, then most probably there was a break into the debugger, so the logging of the message can be skipped. If the exception was caught within fewer than 10 milliseconds, then the debugger probably did not break, so you can proceed with logging the message.
It goes without saying that besides simply logging a message you can try to take as many additional measures as you can, like produce a beep, turn on a lamp, send an email to yourself, bang a gong, fry a motherboard, whatever it takes.
I have a piece of code that looks like this:
Algorithm a = null;
while(a == null)
{
a = grid.getAlgorithm();
}
getAlgorithm() in my Grid class returns some subtype of Algorithm depending on what the user chooses from some options.
My problem is that even after an algorithm is selected, the loop never terminates. However, that's not the tricky bit, if I simply place a System.out.println("Got here"); after my call to getAlgorithm(), the program runs perfectly fine and the loop terminates as intended.
My question is: why does adding that magic print statement suddenly make the loop terminate?
Moreover, this issue first came up when I started using my new laptop, I doubt that's related, but I figured it would be worth mentioning.
Edit: The program in question is NOT multithreaded. The code for getAlgorithm() is:
public Algorithm getAlgorithm ()
{
return algorithm;
}
Where algorithm is initially null, but will change value upon some user input.
I believe the issue has to deal with how grid.getAlgorithm is executed. If there is very little cost associated with executing the method, then your while loop will cycle very quickly as long the method continues to return null. That is often referred to as a busy wait.
Now it sounds like your new laptop is encountering a starvation issue which didn't manifest on your old computer. It is hard to say why but if you look at the link I included above, the Wikipedia article does indicate that busy waits do have unpredictable behavior. Maybe your old computer handles user IO better than your new laptop. Regardless, on your new laptop, that loop is taking resources away from whatever is handling your user IO hence it is starving the process that is responsible for breaking the loop.
You are doing active polling. This is a bad practice. You should at least let the polling thread sleep (with Thread.sleep). Since println does some io, it probably does just that. If your app is not multithreaded it is unlikely to work at all.
If this loop is to wait for user input in a GUI then ouch. Bad, bad idea and even with Thread.sleep() added I'd never recommend it. Instead, you most likely want to register an event listener on the component in question, and only have the validation code fire off when the contents change.
It's more than likely you're program is locking up because you've reached some form of deadlock more than anything else, especially if your application is multithreaded. Rather than try to solve this issue and hack your way round it, I'd seriously consider redesigning how this part of the application works.
You should check getAlgorithm(), there must be something wrong in the method.
There are two scenarios:
Your code is really not meant to be multi-threaded. In this case you need to insert some sort of user input in the loop. Otherwise you might as well leave it as Algorithm a = grid.getAlgorithm(); and prevent the infinite loop.
Your code is multi-threaded in which case you have some sort of 'visibility' problem. Go to Atomicity, Visibility and Ordering or read Java Concurrency in Practice to learn more about visibility. Essentially it means that without some sort of synchronization between threads, the thread you are looping in may never find out that the value has changed due to optimizations the JVM may perform.
You did not mention any context around how this code is run. If it is a console based application and you started from a 'main' function, you would know if there was multi-threading. I am assuming this is not the case since you say there is no multithreading. Another option would be that this is a swing application in which case you should read Multithreaded Swing Applications. It might be a web application in which case a similar case to swing might apply.
In any case you could always debug the application to see which thread is writing to the 'algorithm' variable, then see which thread is reading from it.
I hope this is helpful. In any case, you may find more help if you give a little more context in your question. Especially for a question with such an intriguing title as 'Weird Java problem, while loop termination'.
I am evaluating user inputs as commands for my application. If the user presses Q, or q, and then hits enter, the application quits and execution terminates.
Is there a proper context, or best practices on how to do that? I do not have any resources to release, or anything like that. Should I just use System.exit(0);? Is there a recommended way to do that?
As my first approach I do something like this:
while (true){
try{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
//Other logic goes here...
if (br.readLine().equalsIgnoreCase("Q")){
System.exit(0);
}
}
catch (IOException ioe) {
System.out.println("IO error trying to read your selection");
}
}
You might as well return up to main() and return from there.
private void loop() {
while (true){
try{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
//other logic goes here...
if(br.readLine().equalsIgnoreCase("Q")){
return; // You're done and you are returning to the caller.
}
}
catch (IOException ioe) {
System.out.println("IO error trying to read your selection");
}
}
}
public static void main(String[] args) {
loop();
}
But if you don't have anything to release, System.exit(0) is fine.
Resources are cleaned up automatically by the OS when the process exits. There are two primary reasons to not just exit() from the middle of the code (this applies to all languages).
There may be some action that needs to be taken before the program ends. For example, you may need to save any open files (i.e. write changes that for performance or other reasons have not been sent to the file yet).
Someone may want to later use your code for some other purpose.
Take for example, the Git version control system. There's several efforts to turn the code into a library instead of a set of stand-alone executables so that it can be efficiently incorporated into other projects. One of the problems (from what I've heared) is that the code sometimes simply terminates instead of tracking and cleaning up the resources it's using. As an executable that's fine, but if it was imported as a library, you don't always want to just terminate the host application because you've finished your little part.
Returning all the way back to Returning out of main() is the cleanest way, of course, but if that's not easy to do, System.exit() is perfectly fine.
It's not directly relevant to your question, but throwing an (unhandled) exception is usually the way to terminate on a fatal condition, since it provides a lot of tracing info to the poor user.
I would like to say:
It doesn't matter, it's entirely up to
you!
And dough it sometimes is true, other times it depends on the program structure and code integrity.
The reasonable exit would be through main method, terminating all the threads.
System.exit forces termination of all the threads in the JVM. The System.exit never returns normally. It should be used for an error.
System.exit() is fine. And I agree with the other answers, returning to main would be the best way to exit - from a maintenance point of view. As your code expands you may forget the little method doing System.exit have to debug to remember it.
However you only need to use exit if you have a script which needs the information about an abnormal termination. Otherwise there is no need.
As I recall a Java program will exit with 0 by default - meaning normal termination. So you'd want to use System.exit(n) for cases where your program terminates due to some error.
Example (just using static methods you'd most likely want to instantiate...):
public static void main(String[] args) {
try {
doStuff();
} catch (SomeRuntimeException e) {
// marching orders! Exit with errorcode
<log the error with sufficient info for debugging>
System.exit(1);
}
}
private static doStuff() {
// doing my thing ...
...
//then some error occurs and I have no other choice but to do a hard exit
throw new SomeRuntimeException( ... some info would be nice ...)
...
return
}
Don't forget that if you perform a System.exit(), you can't easily later use your methods in a standalone library. If you want to reuse your code outside your existing solution (and you may not now, but decide to do so in the future), then you'll then have to revisit and refactor appropriately.
Otherwise the worst-case scenario is that you take the existing code without noticing the System.exit()s, use it and then have your library exit unexpectedly in a new application.
Just something to bear in mind re. code-reuse.
I would like to save the programs settings every time the user exits the program. So I need a way to call a function when the user quits the program. How do I do that?
I am using Java 1.5.
You can add a shutdown hook to your application by doing the following:
Runtime.getRuntime().addShutdownHook(new Thread(new Runnable() {
public void run() {
// what you want to do
}
}));
This is basically equivalent to having a try {} finally {} block around your entire program, and basically encompasses what's in the finally block.
Please note the caveats though!
Adding a shutdown hook addShutdownHook(java.lang.Thread) is probably what you look for. There are problems with that approach, though:
you will lose the changes if the program aborts in an uncontrolled way (i.e. if it is killed)
you will lose the changes if there are errors (permission denied, disk full, network errors)
So it might be better to save settings immediately (possibly in an extra thread, to avoid waiting times).
Are you creating a stand alone GUI app (i.e. Swing)?
If so, you should consider how you are providing options to your users how to exit the application.
Namely, if there is going to be a File menu, I would expect that there will be an "Exit" menu item.
Also, if the user closes the last window in the app, I would also expect it to exit the application.
In both cases, it should call code that handles saving the user's preferences.
Using Runtime.getRuntime().addShutdownHook() is certainly a way to do this - but if you are writing Swing applications, I strongly recommend that you take a look at JSR 296 (Swing Application Framework)
Here's a good article on the basics: http://java.sun.com/developer/technicalArticles/javase/swingappfr/.
The JSR reference implementation provides the kind of features that you are looking for at a higher level of abstraction than adding shutdown hooks.
Here is the reference implementation: https://appframework.dev.java.net/