passing variables into methods

[maggie karve]

class Mixer{
Mixer(){
}
Mixer(Mixer m){
m1=m;
}
Mixer m1;
public static void main(String [] args){
Mixer m2=new Mixer();
Mixer m3=new Mixer(m2);
m3.go();
Mixer m4=m3.m1;
m4.go();
Mixer m5=m2.m1;
m5.go();

}
void go(){
System.out.println("hi");
}

how is the heiararchy..how does code get executed???

thanks

[Raju Champaklal]

see if this helps you https://coderanch.com/t/477077/Programmer-Certification-SCJP/certification/Passing-Variables-into-Methods-concept

[maggie karve]

no i have not understood anything from that post...please make it simpler......


thanks

[W. Joe Smith]

My first step would be to take it out of that pretty ugly form. When I see a ; after a method call, I would hit enter to move the next call to the next line. From there just go line by line to see what the code is doing. Since this appears to be about object creation/assignment, creating an object diagram of some type showing what reference is pointing to what object would be a good thing.

[dimple bav]

Hi,

Its not UGLY.

May be some people have different understanding of interpretation of the code line by line.

So thank you.

[W. Joe Smith]

dimple bav wrote:Hi,

Its not UGLY.

May be some people have different understanding of interpretation of the code line by line.

So thank you.

I apologize if my comment offended anyone, but to me if a piece of code has a statement termination it is much easier to start the next statement on the next line. I find this hard to read:

String a = new String("13");Integer b = parseInt(a);int c = b + b;System.out.println(c);
    

But, broken out by statement, it is much simpler (and I find it easier to find errors):

String a = new String("13");
Integer b = parseInt(a);
int c = b + b;
System.out.println(c);
    

[rushikesh sawant]

two Mixer objects are created and each Mixer is having one reference variable, m1.

m2--> 1st Mixer Object
m3--> 2nd Mixer Object

when,
Mixer m3=new Mixer(m2);

executes, m1 in 2nd Mixer is assigned a reference of 1st Mixer object . This happens in 2nd Mixer's one-arg constructor.

so, it is m1-->1st Mixer

m3.go(); calls go method on 2nd Mixer object.

Mixer m4=m3.m1;

This assigns m4 to 1st Mixer. because, m3.m1--->1st Mixer. (m3 is uses m1 in 2nd Mixer, since, m3-->2nd Mixer)
so it is m4-->1st Mixer.

m4.go(); Calls go method on 1st Mixer object.

Mixer m5=m2.m1;

This assigns m5 to null. Because m1 in 1st Mixer object is not referring to any Mixer. so it gets default value, which is null.
so,

m5.go(); results in NullPointerException at runtime.

[dimple bav]

Thank you for such a good interpretation of the code.

[dimple bav]

Its Ok.

I was just trying to interpret each line of code so jumble with the code and comments.

Anyway thank you for the reply.

Dimple.

[maggie karve]

thanks a lot rushikesh!!!you are great!! !

[Bert Bates]

heads up guys -

get used to that UGLY code for the real exam - sometimes they'll squeeze a couple of lines of code onto one line!

[Raju Champaklal]

but i found everything clean in the exam....stop scaring people Bert

[Charles Chikito]

//m3.go() is valid so "hi" is printed
//m4 is referring to m3.m1, m3.m1 is referring to m2, which is also valid
//so "hi is printed for m4.go()
//m5 is referring to m2.m1 which has been initialized with default value, so
//NullPointerException is thrown when m5.go() is invoked


HTH

_charles

[Bert Bates]

Hi Raju,

Just remember, you haven't seen all the questions in Sun's question pool.

[Raju Champaklal]

yeah true

[W. Joe Smith]

Bert Bates wrote:heads up guys -

get used to that UGLY code for the real exam - sometimes they'll squeeze a couple of lines of code onto one line!

Yeah, those tricky exam writers!