Passing Data among Methods

[Prathyu Krishna]

Can someone explain why it doesn't output SPARKY?

public static void main(String args[])
{ 
String name="Webby";
speak(name);
System.out.println(name);
}

public static void speak(String name)
{
name="Sparky";
}

[João Victor Gomes]

Java uses pass-by-value. So, when passing a reference variable, you are passing the value of the reference to the object.
The name parameter of the speak() method receives the copy of the reference, so it points to the same String as the variable name inside the main() method.
Now, remember, Strings are immutable, so you can't change the object's value.
Inside your method speak(), you are assigning a new String to your parameter called name, so it will just point to a different String. The name variable inside the main method still points to "Webby".
Thus, the caller is not affected.

public static void main(String args[]) {
    String name = "Webby"; //1 - creates a String variable, and assign "Webby" to it (a String literal)
    speak(name); //2 - passes the parameter to the method, which is the reference to "Webby"
    System.out.println(name); //5 - prints Webby
}

public static void speak(String name) { //3 - receives the reference to "Webby", so, now it points to "Webby" too
    name = "Sparky"; //4 - now, your name parameter is pointing to another String, rather than "Webby", 
                     //but the name variable of the main method still points to "Webby"
}

[Prathyu Krishna]

See this example..

My question is:
1. Strings are immutable then why does it append "d" ?
2. totally confused what reference variables are?

public class ReturningValues{
public static void main(String args[]){
int number=1;
String letters="abc";
number(number);
letters =  letters(letters);
System.out.println(number+letters); //1abcd
}


public static int number(in number){
number++;
return number;
}

public static String letters(String letters){
letters+="d";
return letters;
}
}

[Roel De Nijs]

Prathyu Krishna wrote:1. Strings are immutable then why does it append "d" ?

Because the statement letters+="d"; is equivalent to

String temp = letters + "d";
letters = temp;

So a new String object "abcd" was created, because strings are indeed immutable!

Prathyu Krishna wrote:2. totally confused what reference variables are?

Are you using a study guide to prepare for this certification exam? If you don't, I strongly advise to purchase and use one during your preparation.

[João Victor Gomes]

Prathyu Krishna wrote:2. totally confused what reference variables are?

A reference variable is a variable that contains a reference to an object in the memory (so, the variable holds/points to an object in the memory, or points to null).
For example, assume that you have a class called Car, we can declare a reference variable of it:

class Car {
	//Code here...
}

class CarTest {
	public static void main(String[] args) {
        //car is a reference variable of type Car. It holds an object of type Car (by assigning new Car() to it).
		Car car = new Car();
	}
}

Prathyu Krishna wrote:1. Strings are immutable then why does it append "d" ?

Be careful, you are not appending "d" to your String. You are doing a concatenation, which in this case creates a new String (as Roel already explained).
Your method letters() is returning the new String created by the concatenation.

public class ReturningValues {
	public static void main(String args[]) {
		int number = 1;
		String letters = "abc";
		number(number);
		letters = letters(letters); //assigns the new returned String from the method to the variable letters.
		System.out.println(number + letters);
	}

	public static int number(in number) {
		number++;
		return number;
	}

	public static String letters(String letters) {
        //concatenation of letters and "d", creating a new String that is assigned to the same letters variable (the parameter)
		letters += "d";
        //returns the reference to the new String created inside this method by the above concatenation 
		return letters;
	}
}

As you see, you are not appending, you are doing a String concatenation, creating a new String, which means that you are not changing the original object (since Strings are immutable).
And be careful with variables and methods names. Search about good practice when naming variables and methods (like not using the same name of a variable to a method)

[Roel De Nijs]

João Victor Gomes wrote://car is a reference variable of type Car. It holds an object of type Car (by assigning new Car() to it).

That's not 100% accurate. It should be: "car is a reference variable of type Car. It holds a reference to an object of type Car (by assigning new Car() to it)".

[João Victor Gomes]

Roel De Nijs wrote:

João Victor Gomes wrote://car is a reference variable of type Car. It holds an object of type Car (by assigning new Car() to it).

That's not 100% accurate. It should be: "car is a reference variable of type Car. It holds a reference to an object of type Car (by assigning new Car() to it)".

Yes, I mentioned that on my explanation

João Victor Gomes wrote:A reference variable is a variable that contains a reference to an object in the memory

But when talking about the car variable, I was just trying to be more informal, but maybe it wasn't a good idea.
Thanks for de reminder.

[Prathyu Krishna]

I got it. Thanks for taking time to explain this so well!

[Roel De Nijs]

João Victor Gomes wrote:But when talking about the car variable, I was just trying to be more informal, but maybe it wasn't a good idea.

In the OCAJP forum there's no room for being informal We have the Meaningless Drivel forum for that