How can I get the memory location of a object in java?

Solution 1:

This is something you probably don't want to do.

If you really want to do this, something like this code might help:

package test;

import java.lang.reflect.Field;

import sun.misc.Unsafe;

public class Addresser
{
    private static Unsafe unsafe;

    static
    {
        try
        {
            Field field = Unsafe.class.getDeclaredField("theUnsafe");
            field.setAccessible(true);
            unsafe = (Unsafe)field.get(null);
        }
        catch (Exception e)
        {
            e.printStackTrace();
        }
    }

    public static long addressOf(Object o)
    throws Exception
    {
        Object[] array = new Object[] {o};

        long baseOffset = unsafe.arrayBaseOffset(Object[].class);
        int addressSize = unsafe.addressSize();
        long objectAddress;
        switch (addressSize)
        {
            case 4:
                objectAddress = unsafe.getInt(array, baseOffset);
                break;
            case 8:
                objectAddress = unsafe.getLong(array, baseOffset);
                break;
            default:
                throw new Error("unsupported address size: " + addressSize);
        }       

        return(objectAddress);
    }


    public static void main(String... args)
    throws Exception
    {   
        Object mine = "Hi there".toCharArray();
        long address = addressOf(mine);
        System.out.println("Addess: " + address);

        //Verify address works - should see the characters in the array in the output
        printBytes(address, 27);

    }

    public static void printBytes(long objectAddress, int num)
    {
        for (long i = 0; i < num; i++)
        {
            int cur = unsafe.getByte(objectAddress + i);
            System.out.print((char)cur);
        }
        System.out.println();
    }
}

But

• not portable across JVMs or even different versions
• objects can move because of GC at any time, and cannot synchronize across GCs so results might not make sense
• not tested across all architectures, endianess, etc. might make this not work everywhere

Solution 2:

Without using JVM-specific features this cannot be done. Java deliberately hides the location associated with each object to give the implementation more flexibility (the JVM often moves objects around in memory when doing garbage collection) and to improve security (you can't use raw pointers to trash memory or access nonexistent objects).