Memory address of variables in Java

Question

That is the class name and System.identityHashCode() separated by the '@' character. What the identity hash code represents is implementation-specific. It often is the initial memory address of the object, but the object can be moved in memory by the VM over time. So (briefly) you can't rely on it being anything.

Getting the memory addresses of variables is meaningless within Java, since the JVM is at liberty to implement objects and move them as it seems fit (your objects may/will move around during garbage collection etc.)

Integer.toBinaryString() will give you an integer in binary form.

It is possible using sun.misc.Unsafe : see this great answer from @Peter Lawrey -> Is there a way to get a reference address?

Using its code for printAddresses() :

    public static void printAddresses(String label, Object... objects) {
    System.out.print(label + ": 0x");
    long last = 0;
    int offset = unsafe.arrayBaseOffset(objects.getClass());
    int scale = unsafe.arrayIndexScale(objects.getClass());
    switch (scale) {
    case 4:
        long factor = is64bit ? 8 : 1;
        final long i1 = (unsafe.getInt(objects, offset) & 0xFFFFFFFFL) * factor;
        System.out.print(Long.toHexString(i1));
        last = i1;
        for (int i = 1; i < objects.length; i++) {
            final long i2 = (unsafe.getInt(objects, offset + i * 4) & 0xFFFFFFFFL) * factor;
            if (i2 > last)
                System.out.print(", +" + Long.toHexString(i2 - last));
            else
                System.out.print(", -" + Long.toHexString( last - i2));
            last = i2;
        }
        break;
    case 8:
        throw new AssertionError("Not supported");
    }
    System.out.println();
}

I set up this test :

    //hashcode
    System.out.println("Hashcode :       "+myObject.hashCode());
    System.out.println("Hashcode :       "+System.identityHashCode(myObject));
    System.out.println("Hashcode (HEX) : "+Integer.toHexString(myObject.hashCode()));

    //toString
    System.out.println("toString :       "+String.valueOf(myObject));

    printAddresses("Address", myObject);

Here is the output :

Hashcode :       125665513
Hashcode :       125665513
Hashcode (HEX) : 77d80e9
toString :       java.lang.Object@77d80e9
Address: 0x7aae62270

Conclusion :

• hashcode != address
• toString = class@HEX(hashcode)

That is the output of Object's "toString()" implementation. If your class overrides toString(), it will print something entirely different.

This is not memory address This is classname@hashcode
Which is the default implementation of Object.toString()

public String toString() {
    return getClass().getName() + "@" + Integer.toHexString(hashCode());
}

where

Class name = full qualified name or absolute name (ie package name followed by class name) hashcode = hexadecimal format (System.identityHashCode(obj) or obj.hashCode() will give you hashcode in decimal format).

Hint:
The confusion root cause is that the default implementation of Object.hashCode() use the internal address of the object into an integer

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.

And of course, some classes can override both default implementations either for toString() or hashCode()

If you need the default implementation value of hashcode() for a object which overriding it,
You can use the following method System.identityHashCode(Object x)