String and StringBuffer

[Gunjaan Agarwal]

How is String passed in method call(ref or value)??
Kindly explain me the below program output


public class Test{

public static void main(String [] args) {
String textString = new String ("java");
StringBuffer textBuffer = new StringBuffer ("java");
stringReplace (textString);
bufferReplace (textBuffer);
System.out.println (textString + textBuffer);
}

public static void stringReplace (String text) {
text = text.replace ('j' , 'i');
//System.out.println(text);
}

public static void bufferReplace (StringBuffer text) {
text = text.append ("C");
//System.out.println(text);

}

}

output is javajavac

[Shiaber Shaam]

In general non primitives are passed by references only.. since string objects are immutable their originak values cant be changed... and string buffer are mutable and hence value gets changed...

Am i right... Do send more Explanations...

[Amit Goyal]

Every thing in java is pass by value. u can check previous discussion on this topic.

[S Thiyanesh]

Java uses only pass by value.
In case of the String, the new Object is created inside the stringReplace but it doesn't affect the original String object in the calling method.
In case of StringBuffer, the object state is modified and hence the change is reflected in the calling method.

[joshua antony]

yes reference is passed, but it is passed by value

[wise owen]

Always Pass By Value in Java.

[Ramasubbhu Allur Kuppusamy]

Java uses pass by value. The value of the reference will be passed in the case of objects.

[S Thiyanesh]

Pass by reference means you should be able to change the value stored in the reference variable also.
But in JAVA only the objects pointed by the reference variable is changed not the reference variable itself.
So in JAVA there is ONLY PASS BY VALUE.