How exactly works the Java application exit code of the main() method?
exit-code, java, program-entry-point
Solution
The VM exits when
- all of the non-daemon threads stop running, or
- `System.exit(exitCode)` is called
In the first case, the exit code is 0. In the second case, it's the exit code passed to the `exit()` method.
Don't forget that even if your main() method returns, the program will continue running until no non-daemon thread runs anymore. And any thread running in the VM can choose to exit explicitely.
The exit code 0 means that everything went as expected. you can choose to use any other exit code to signal an exceptional condition to the environment.
Problem
I have the following doubts related a simple command line Java application. So I have this command line application that is started by a `main()` method defined inside a Main class. As usual this `main()` method is defined with this signature: ``` public static void main(String[] args) { ``` It's return type is `void`, and that should mean it doesn't return any value. But when its execution correctly terminates I obtain following message in the IntelliJ console. ``` Disconnected from the target VM, address: '127.0.0.1:54090', transport: 'socket' Process finished with exit code 0 ``` What exactly does represent the `exit code 0`? I think it means that the program have correctly completed its execution without incur into any error. So now I have the following 2 doubts: If it is true why it happens if my `main()` method return `void`? How can I return a different exit code if my application ended with an error? Is there a standard exit code value for ending with errors?