Confusion on how Java StringBuffer works?

Question

I'm working on an assignment right now, and am confused by this. The code given is simple:

public class Variables{
    public static void main(String[ ] args){
        StringBuffer b = "ghi";
        f(b);
        System.out.println(b):
    }

    public static void f(StringBuffer p){
        p.concat("jkl");
    }
}

The question simply asks what the output of the print statement will be. My first thought was simply "ghi", but this was incorrect. If the method f is taking b as a parameter and setting it to p, then how does .concat() modify b? I've read through the StringBuffer documentation and dont understand why this wouldn't end up with b equaling "ghi" while p is "ghijkl".

Basically, how is the .concat() method called on p also modifying b?

I doesn't modify `b`, it modifies the object *referenced* by `b`. `b` is *not* a StringBuffer, it's a *reference* (pointer) to a StringBuffer. (Understanding the difference between a *reference* and the thing *referenced* is critical to understanding most languages more sophisticated than BASIC.) — Hot Licks, Oct 15 '14 at 01:28

score 2 · Answer 1 · edited May 23 '17 at 10:24

2

First, you need to create a StringBuffer correctly.

StringBuffer foo = new StringBuffer("some string");

You passed an object to a function. Java passes objects as references (the references themselves of course are passed by value). See: Is Java "pass-by-reference" or "pass-by-value"?

Since you have a reference to a StringBuffer and not a copy of it, you are actually modifying the same object as you would as if you were in main.

edited May 23 '17 at 10:24

Community

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answered Oct 15 '14 at 01:32

csnate

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I can't test it now, but I'm not sure you can assign a StringBuffer in that way. Don't you have to do `StringBuffer foo = new StringBuffer("some string here")`? – csnate Oct 15 '14 at 01:34
you are right it is the wrong way. please add to your answer so I can vote u up – Kick Buttowski Oct 15 '14 at 01:35

score 1 · Answer 2 · answered Oct 15 '14 at 01:33

This is the difference between parsing by reference and parsing by value.

When you create a method taking a primitive, for example, it undergoes what is called parsing by value. The JVM pretty much creates a copy of the value and that is what you get in the method.

int j = 0;
foo(k);
System.out.println(j); //Will still be 0.

public static void foo(int i) {
 //i is not j (what you called the method with), it is a copy of it that is only valid within this method
 i++;
}

Now when you call the method with an object such as a StringBuffer you are parsing by reference. This means that you are not parsing the value of the StringBuffer(not parsing a StringBuffer` with "ghi" as its contents), but you are rather parsing a pointer to it. Using this pointer you can still operate on the original object within other methods and bodies of code.

Confusion on how Java StringBuffer works?

2 Answers2