Understanding syntax for running Java program in command line

Question

I'm trying to understand this batch job file, there are two and the first is returning an exit code of 0:

set JAVA_HOME="C:\Program Files(x86)\Java\jre1.8.0_221\bin"

%JAVA_HOME%\java -Xms125M -Xmx512M -Djava.ext.dirs=lib org.pg.test.ListOutput > output.txt 2>exception.txt

And this second one is returning an exit code of 1:

set JAVA_HOME="C:\Program Files(x86)\Java\jre1.8.0_221\bin"

%JAVA_HOME%\java -Xms125M -Xmx512M -Djava.ext.dirs=lib org.pg.test.ListOutput 1 0 > output.txt 2>exception.txt

What does the 1 & 0 after the file name mean exactly? Why is this making it return an exit code 1?

Both files run normally, with a successful output in output.txt.

Trying to google the exact syntax this is in (batch files were not written by me) but maybe I'm searching for the wrong thing.

Answer 1

The documentation for the java command is found

• here for JDK/JRE version 18 (a fairly recent version), and
• here for JDK/JRE version 1.8 for Windows, which appears to be what you're using.

The 1 and the 0 after the class name are arguments that will be passed to the Java program when it runs. As for why this causes the program to exit with a particular exit code, that depends on what the program does. It's impossible to answer that without seeing the source code of the program.