Behaviour of String as reference type

Question

I tried this simple code to find out how a String behaves. It does not work as you could expect from a reference type. On the other hand the keyword new is leftout but I expect it is used in the background.
Can anyone explain !
Thanks.
class Te {
public static void main (String arg[]){
     int[] a = new int [1];
     a[0] = 1;
     int[] b = a;
     a[0] = 2;
     System.out.println("a[0]="+a[0]+" b[0]="+b[0]);
     String c = "AA";
     String d = c;
     c = "BB";
     System.out.println("c="+c+" d="+d);
And the printout is:
a[0]=2 b[0]=2
c=BB d=AA

Answer 1

I I understand it correctly, the line
c = "BB";is realy
c = new String("BB");otherwise c and d should have reference to the same
String.On some level, yes, but in reality...
If you know C++, it is should be enough just to say that java's references are really pointers - there are also references in C++ but they mean another things.
So, c = "BB" just assigns the "reference variable" (pointer) c to a new reference of a String object, but the value of the "reference variable" (pointer) d remains the same. So basicallyString c = "AA";
String d = c;
c = "BB";is the same thing asint c = 1;
int d = 2;
c = 1;Strings in Java have this nice little extra feature... saying that String c = "BB" and String c = new String("BB") are totally different things. You can check this with code like:String a = "xyzzy";
String b = "xyzzy";
String c = new String("xyzzy");
if (a == b) System.out.println("a and b refer to the same object");
if (a != c) System.out.println("a and c don't refer to the same object");The literal Strings are stored in a literal String pool, so each occurence of "BB" refers to the same object. But new String("BB") creates a whole new String object, different from the literal "BB".