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Are Objects passed by value or reference?

 
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I found this one of the mock exams . It is quite interesting Please follow the code below.
public class StrBufTest {
public void method1(StringBuffer s1, StringBuffer s2) {
s1.append("There");
s2 = s1; //line 4
}
public static void main(String args[]) {
StringBuffer sb1 = new StringBuffer("Hello");
StringBuffer sb2 = new StringBuffer("Hello");
StrBufTest sbt = new StrBufTest();
sbt.method1(sb1, sb2);
System.out.println("sb1 is " + sb1 + "\nsb2 is " + sb2);
}
}
The Answer to the above question is
sb1 is "Hello There"
sb2 is "Hello".
I can't understand exactly why the value of sb2 is "Hello" and not "Hello There" since at line 4 s2 references the same object as s1. My question is are objects passed by reference. If they are then sb2 should be "Hello There". If they r passed by value then the answer is justified. I am confused because I have read Khalid's book and it mentions that objects are passed by reference.
Please fill me on the same.

 
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The object is passed by reference, but the object reference variable is passed by value. When you pass an object reference variable to a function, you can potentially change the state of the object, but you can't make the passed reference variable refer to another object. An example should help here:
class Apple { boolean yummy; }
...
Apple a1 = new Apple();
tryChanger(a1); //pass copy of a1's reference
System.out.println("Apple is yummy? " + a1.y); //fine!
...
void tryChanger(Apple inA) //See as "Apple inA = a1;"
{
inA.yummy = true; //I can change its state! Affects inA and a1
inA = null; //this only affects inA!
}
...
As for the variables a1 and inA look at it like this...
Apple a1 = new Apple();
Apple inA = a1; //inA now alias of a1
inA.yummy = true; //what you do to inA affects a1
inA = null; //but a1 still has the object
System.out.println("Apple is yummy? " + a1.y);
Hope this helps
 
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Originally posted by Amit Punjwani:
I found this one of the mock exams . It is quite interesting Please follow the code below.
public class StrBufTest {
public void method1(StringBuffer s1, StringBuffer s2) {
s1.append("There");
s2 = s1; //line 4
}
public static void main(String args[]) {
StringBuffer sb1 = new StringBuffer("Hello");
StringBuffer sb2 = new StringBuffer("Hello");
StrBufTest sbt = new StrBufTest();
sbt.method1(sb1, sb2);
System.out.println("sb1 is " + sb1 + "\nsb2 is " + sb2);
}
}
The Answer to the above question is
sb1 is "Hello There"
sb2 is "Hello".
I can't understand exactly why the value of sb2 is "Hello" and not "Hello There" since at line 4 s2 references the same object as s1. My question is are objects passed by reference. If they are then sb2 should be "Hello There". If they r passed by value then the answer is justified. I am confused because I have read Khalid's book and it mentions that objects are passed by reference object sb2 is unchnged.
Please fill me on the same.


I believe function´┐Żs arguments in Java are passed by value. So in the case of objects it is copy of the reference to the object. Copies of references to the sb2 and sb2 are put on the stack then the call to your function is made. When you assignee value of s1 to s2 all you do is change the copy which has no effect on the object referenced by it or the reference to the object in the calling code, therefore when the functions returns object referenced by sb2 is unchanged.
 
Anonymous
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Hi Amit!
DaB would have cleared your doubt, it not, refer to page 79 in your Kalid Azim Mughal and Rolf Rasmussen's book. In the middle of the page he already explained this topic with reference to previous examples (3.2 & 3.3). The paragraph starts like this "The parameter passing strategy in Java is call-by-value and not call-by-reference, regardless of the type of the parameter. .... ... "
Bye!
Kumaresan.
 
Amit Punjwani
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Thanx guys I got it. I was stuck on it for quite sometime.

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