can someone explain this behavior of StringBuffer? - stringbuffer

Question

public class StrBuffer {
public static void main(String[] args) {
StringBuffer sb = new StringBuffer(); //5
sb.append("hello"); //6
foo(sb); //7
System.out.println(sb); //8
}
private static void foo(StringBuffer sb) {
// TODO Auto-generated method stub
sb.append("wow"); //1
sb = new StringBuffer(); //2
sb.append("foo"); //3
System.out.println(sb); //4
}
}
in the above when i print in line8. output is "hellowow" .... can some one explain this please?

Answer 1

sb.append("wow"); //1
You just mutated the StringBuffer instance you passed in.
sb = new StringBuffer(); //2
You just assigned your local parameter to point to a new StringBuffer instance. This has no effect on the caller or on the old instance.

Answer 2

At first, you pass a reference to main's sb to foo.
In foo, you append "wow" to the referenced object.
Then, in foo, you create a new StringBuffer object, and change the local sb to point to it.
It might be clearer if I rename the reference:
public static void main(String[] args) {
StringBuffer sb = new StringBuffer(); //5
sb.append("hello"); //6
foo(sb); //7
System.out.println(sb); //8
}
private static void foo(StringBuffer ref) {
ref.append("wow"); //1
ref = new StringBuffer(); //2
ref.append("foo"); //3
System.out.println(ref); //4
}
'ref', as passed in to foo, is a just a reference to main's sb. You can operate on ref (ie., sb) in foo until you changed ref to refer to a different object.