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Java Programe

 
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Hi All,

Bellow i'm writing one small programe,


public class TrickyOne {
TrickyOne to = new TrickyOne("welcome");

TrickyOne(String a){
this.localvariable = a;
}

private String localvariable;

/**
* @return the localvariable
*/
public String getLocalvariable() {
return localvariable;
}

/**
* @param localvariable the localvariable to set
*/
public void setLocalvariable(String localvariable) {
this.localvariable = localvariable;
}

public static void main(String args[]) {
TrickyOne to = new TrickyOne("welcome");
System.out.println(to);
}
}

When your displaying 'TrickyOne object(to) ,Is it possible to display welcome.Should not do any chnages inside main method.


Thanks & Regards,
Sijesh
 
author and iconoclast
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When you print an object using println(), the object's toString() method is called. So give this class a toString() method which returns what you want printed.

I should point out that the variable you've named "localvariable" is not a local variable -- it's a member variable. A local variable is one defined inside a method.
 
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Dear Sijesh Alayaril
It is my suggestion to use code tags every time when you post a code so that we can read it without any difficulties.
 
Sijesh Alayaril
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Thank You Ernest Friedman-Hill .
 
Sijesh Alayaril
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Hi Nikhil Sagar ,

Next time onwards i will follow the code tags.

Thanks & Regards,
Sijesh
 
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Sijesh Alayaril wrote:


Also, your code can throw StackOverflowError.. Check your instance reference variable... it will be initialized for every instance of TrickyOne you create.. Which in turn again create an instance of TrickyOne..

It is a chain of instances like: - TrickyOne contains(TrickyOne contains(TrickyOne(contains TrickyOne(..... so on..))))
Beware of these kinds of designs..
 
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R. Jain wrote:

Sijesh Alayaril wrote:


Also, your code can throw StackOverflowError.. Check your instance reference variable... it will be initialized for every instance of TrickyOne you create.. Which in turn again create an instance of TrickyOne..

It is a chain of instances like: - TrickyOne contains(TrickyOne contains(TrickyOne(contains TrickyOne(..... so on..))))
Beware of these kinds of designs..



there is no CAN. it will definitely throw StackOverflowError.

the other day i realised that if you are reading JLS the words CAN , MAY sometimes makes a huge difference. so i thought to correct you although i know you intended that it WILL throw error for sure. Regards
 
R. Jain
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gurpeet singh wrote:so i thought to correct you although i know you intended that it WILL throw error for sure. Regards


Oh Yeah.. Words do matter.
Correction appreciated.. and Accepted
 
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