Executing a Java application in a separate process

Can a Java application be loaded in a separate process using its name, as opposed to its location, in a platform independent manner?

I know you can execute a program via ...

Process process = Runtime.getRuntime().exec( COMMAND );

... the main issue of this method is that such calls are then platform specific.

Ideally, I'd wrap a method into something as simple as...

EXECUTE.application( CLASS_TO_BE_EXECUTED );

... and pass in the fully qualified name of an application class as CLASS_TO_BE_EXECUTED.

This is a synthesis of some of the other answers that have been provided. The Java system properties provide enough information to come up with the path to the java command and the classpath in what, I think, is a platform independent way.

public final class JavaProcess {

    private JavaProcess() {}        

    public static int exec(Class klass, List<String> args) throws IOException,
                                               InterruptedException {
        String javaHome = System.getProperty("java.home");
        String javaBin = javaHome +
                File.separator + "bin" +
                File.separator + "java";
        String classpath = System.getProperty("java.class.path");
        String className = klass.getName();

        List<String> command = new LinkedList<String>();
        command.add(javaBin);
        command.add("-cp");
        command.add(classpath);
        command.add(className);
        if (args != null) {
            command.addAll(args);
        }

        ProcessBuilder builder = new ProcessBuilder(command);

        Process process = builder.inheritIO().start();
        process.waitFor();
        return process.exitValue();
    }

}

You would run this method like so:

int status = JavaProcess.exec(MyClass.class, args);

I thought it made sense to pass in the actual class rather than the String representation of the name since the class has to be in the classpath anyways for this to work.

Two hints:

System.getProperty("java.home") + "/bin/java" gives you a path to the java executable.

((URLClassLoader) Thread.currentThread().getContextClassLoader()).getURL() helps you to reconstruct the classpath of current application.

Then your EXECUTE.application is just (pseudocode):

Process.exec(javaExecutable, "-classpath", urls.join(":"), CLASS_TO_BE_EXECUTED)

Expanding on @stepancheg's answer the actual code would look like so (in the form of a test).

import org.junit.Test;

import java.io.File;
import java.net.URL;
import java.net.URLClassLoader;
import java.util.Arrays;
import java.util.stream.Collectors;

public class SpinningUpAJvmTest {
    @Test
    public void shouldRunAJvm() throws Exception {
        String classpath = Arrays.stream(((URLClassLoader) Thread.currentThread().getContextClassLoader()).getURLs())
                .map(URL::getFile)
                .collect(Collectors.joining(File.pathSeparator));
        Process process = new ProcessBuilder(
                System.getProperty("java.home") + "/bin/java",
                "-classpath",
                classpath,
                MyMainClass.class.getName()
                // main class arguments go here
        )
                .inheritIO()
                .start();
        int exitCode = process.waitFor();
        System.out.println("process stopped with exitCode " + exitCode);
    }
}