How do you get the "object reference" of an object in java when toString() and hashCode() have been overridden?

What exactly are you planning on doing with it (what you want to do makes a difference with what you will need to call).

hashCode, as defined in the JavaDocs, says:

As much as is reasonably practical, the hashCode method defined by class Object does return distinct integers for distinct objects. (This is typically implemented by converting the internal address of the object into an integer, but this implementation technique is not required by the Java™ programming language.)

So if you are using hashCode() to find out if it is a unique object in memory that isn't a good way to do it.

System.identityHashCode does the following:

Returns the same hash code for the given object as would be returned by the default method hashCode(), whether or not the given object's class overrides hashCode(). The hash code for the null reference is zero.

Which, for what you are doing, sounds like what you want... but what you want to do might not be safe depending on how the library is implemented.

This is how I solved it:

Integer.toHexString(System.identityHashCode(object));

Double equals == will always check based on object identity, regardless of the objects' implementation of hashCode or equals. Of course - make sure the object references you are comparing are volatile (in a 1.5+ JVM).

If you really must have the original Object toString result (although it's not the best solution for your example use-case), the Commons Lang library has a method ObjectUtils.identityToString(Object) that will do what you want. From the JavaDoc:

public static java.lang.String identityToString(java.lang.Object object)

Gets the toString that would be produced by Object if a class did not override toString itself. null will return null.

 ObjectUtils.identityToString(null)         = null
 ObjectUtils.identityToString("")           = "java.lang.String@1e23"
 ObjectUtils.identityToString(Boolean.TRUE) = "java.lang.Boolean@7fa"