Question

Please pardon the beginner question.I met this code fragment on the internet:

public class Person {

public static void main(String [] args)
{
    StringBuffer a = new StringBuffer("A");
    StringBuffer b = new StringBuffer("B");
    operate(a,b);
    System.out.println(a+","+b);
}


static void operate(StringBuffer x, StringBuffer y)
{
    y.append(x);
    y=x;

 }
}

I figured out the running output should be A,A, however, the correct output should be A,BA, could expert help me understand why the value of b is still AB? Why “y.append(x)” will affect the value of b, but not “y=x”? That is where I am getting confused.

Thanks in advance.

Answer 1

The short answer is that Java arguments are passed by value, so the final assignment in operate doesn’t have any effect.

The detailed sequence of events is as follows:

1. a and b are initialized to two StringBuffer objects. Lets write this as

    a -> S1{"A"}
    b -> S2("B")
   

2. The (trivial) argument expressions for the operator call arguments are evaluated giving S1{“A”} and S2{“B”}.

3. The call starts, and the two references are assigned to the local variables x and y. So the state is now:

    a -> S1{"A"}
    x -> |
   
    b -> S2("B")
    y -> |
   

4. The y.append(x) call modifies the S2 object:

    a -> S1{"A"}
    x -> |
   
    b -> S2("BA")
    y -> |
   

5. The y = x; assignment is performed:

    a -> S1{"A"}
    x -> |
    y -> |
   
    b -> S2("BA")
   

6. The operate method returns, causing x and y to go out of scope.

    a -> S1{"A"}
    b -> S2("BA")
   

The key thing to note is that in step 5 we DID NOT change the contents of the S2 object. Instead, we just changed y to refer to the other object.