stringbuffers eg from overalltesting.com

[JayaSiji Gopal]

  
 
class ClassA { 
    public static void changeBuffers(StringBuffer x, StringBuffer y) { 
        x.append(y); 
        y = x; 
    } 
 
    public static void main(String[] args) { 
        StringBuffer a = new StringBuffer("A"); 
        StringBuffer b = new StringBuffer("B"); 
        changeBuffers(a, b); 
        System.out.println(a + "," +b); 
    } 
} 
 
 
Question   
 
What is the result of attempting to compile and run the program? 
 
Answers   
 
Choice 1   A,B 
Choice 2   AB,B 
Choice 3   A,AB 
Choice 4   AB,AB 
Choice 5   Code will not compile 
 

I answered 4, but the correct choice is 2. In the statement in the code aboce, we have y=x. ie all tht is in x, ie AB is stored in B. Also objects are passed by the reference. so the original object gets affected. Hence, i answered 4. please correct me.

[Martin Mathis]

I believe this exact question was discussed here recently. Basically what it comes down to is, when you pass a reference into a function, it is actually sending in a COPY of that reference. So there are two unique references pointing to the same thing.

Thus, when you pass a and b into this function, the x.append actually changes A, because a and x are both pointing to the same thing. b remains unchanged because all the function is doing is taking the reference y (which pointed to the same thing that b points to) and points it to x. This causes no change in what b itself points to.