StringBuffer behaviour

Question

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

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

    }
}

Ans is AB,B Please tell me why value of b is unchanged.

http://stackoverflow.com/questions/40480/is-java-pass-by-reference — assylias, Aug 27 '13 at 12:41

score 1 · Answer 1 · answered Aug 27 '13 at 12:47

1

You are passing values of the variables x and y, not a reference to them. So the changed made inside the operate function are useless, follow the link provided as a comment to your post to understand better ;)

answered Aug 27 '13 at 12:47

Alessio

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score 0 · Answer 2 · answered Aug 27 '13 at 12:48

Passing StringBuffer in parameter of method operate is passbyvalue. In this type, the value of variable is just copied into the temp variable defined in method's parameter. Means in your example value of both a and b are just copied in x and y. Here value of a got changed but b didn't.
Its obvious. Here value of b didnt change because y is the mirror copy of b. But append function works with actual value. It takes reference of x and append the y with it and ultimately it shows change at a.

score 0 · Answer 3 · answered Aug 27 '13 at 12:50

0

This is because in Java arguments are passed by value. So y = x just changes local var y value in operate method.

answered Aug 27 '13 at 12:50

Evgeniy Dorofeev

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StringBuffer behaviour

3 Answers3