Why if (arg0.length > 0 && arg0[0].equals(“exitcode”))?

问题

I found this SpringBoot Application code which throws Exit Code 10 when executed with argument "exitcode".

public class Swagger2SpringBoot implements CommandLineRunner {

    @Override
    public void run(String... arg0) throws Exception {
        if (arg0.length > 0 && arg0[0].equals("exitcode")) {
            throw new ExitException();
        }
    }

    public static void main(String[] args) throws Exception {
        new SpringApplication(SpringBootEntityApp.class).run(args);
    }

    class ExitException extends RuntimeException implements ExitCodeGenerator {
        private static final long serialVersionUID = 1L;

        @Override
        public int getExitCode() {
            return 10;
        }

    }

}

Running this code with the given argument exits the program.

java -jar exitcode arg1 arg2 

I would be grateful if somebody could explain the use-case of this code.

PS: Why are we running the program to exit the program.

References:

1. Swagger Petstore.

2. Plaza Client

3. Yidi

4. Stack Overflow Question 1

5. Docker Dash

来源：https://stackoverflow.com/questions/58618988/why-if-arg0-length-0-arg00-equalsexitcode