sakthi moorthy

HI Guys,

I need suggestion for the following scenario
1) There are X number of word Documents saved in a hard disk
2) In Web application upon some action
2.1) the logged in user details has to be fetched from database
2.2) the User details(userId,UserName,Desigantion (hardly some 9 informations)) has to be filled in one of the documents, which has been mentioned in the Step 1.
3) Generate the PDF with the modified content and send it to browser for download.

Solutions
1) Load All the documents into memory as a String or StringBuffer format while application server start up & replace(fill userid, userName and some information) the String or String Buffer & Generate the PDF
2) Load only the requested or specific document as a String or StringBuffer format & replace the String or String Buffer & Generate the PDF
3) Take the contents of word as constants of String Objects and create java program for generating the PDF.


I Think the first two approach will give OutOfMemoryException if everything gets loaded into memory, The Third approach will be performance impact because of more String operations.

how to do it in efficient way.


Thanks & Regards
S.Sathiya

Wow Thanks SeethaRaman Venkatasamy, gurpeet singh. and while reading JLS, any suggestions or steps to be taken care, before reading and understanding some complex stuff's in Java Language Specification, because i do not want to grasp the concepts wrongly.

Hi nam ha minh

yeah its my mistake making the outter class proctected, i have created an inner class and given public, protected access specifiers, still compilation fails.

public class JLS811ClassModifiers { public protected class InnerClass{ } }

Hi ,

The following sentences have been taken from the Java Language Specification Third Edition. and Section 8.1.1

A class declaration may include class modifiers.
ClassModifiers:
ClassModifier
ClassModifiers ClassModifier
ClassModifier: one of
Annotation public protected private
abstract static final strictfp

........................
..................
......................


If two or more class modifiers appear in a class declaration, then it is customary,
though not required, that they appear in the order consistent with that shown
above in the production for ClassModifier.

so from the above quote i understand that a class can have two or more class modifiers and I have tried to create a java class with two modifiers as following

public protected class JLS811ClassModifiers { }


i am getting compilation error. or is there wrong in my understanding.

Thanks Stuart i got it now


I wanted to convey one more thing here the code
public class HHH { private boolean s = false; { // Instance Intializer block if (s) throw new IOException(); // line1 } HHH() throws IOException { // explictly declaring the IOException in throws caluse of constructor && i have atleast one default Constructor too } }

will compile fine until the default Constructor HHH() throws IOException or IOException's super type, which means if you want to throw a Exception in Intializer block, the same Exception should be thrown in all Constructors thats what the JLS Says. so the following code wont compile.



public class HHH { boolean s = false; { if(s) throw new IOException(); // compile time error here..... } HHH(){ } }


Again thanks Stuart

Hi Stuart,


Thanks for the reply. I will tell my understanding of the quoted JLS Sentence.

It is compile-time error if an instance initializer of a named class can throw a
checked exception unless that exception or one of its supertypes is explicitly
declared in the throws clause of each constructor of its class and the class has at
least one explicitly declared constructor

In order to throw a checked exception from Instance Intializer following conditions needs to be statisfied

1) At Least One explicitly declared Constructor Say HHH() in this case has to be present
2) The Same that is HHH() Constructor has to throws A Exception or the Exceptions Super Type, so declared the constructor like following
HHH() throws IOException { }


If above conditions are satisfied, you can have an instance intializer block that can throw a Checked Exception like following
{ throw new IOException(); // line1 }

I am not able to relate my understanding with your code, to be specific i am not able to understand why you have kept if block in instance intializer. If i am not wrong, statements under conditional checking block will be executed only on certain condition or conditions, which may or may not happen,thats why that compilation problem has resolved.

Hi All,

I am refering the JLS (Instance Intializers Chapter 8.6) which tells the following thing

An instance initializer declared in a class is executed when an instance of the class
is created (§15.9), as specified in §8.8.7.1.
InstanceInitializer:
Block

It is compile-time error if an instance initializer of a named class can throw a
checked exception unless that exception or one of its supertypes is explicitly
declared in the throws clause of each constructor of its class and the class has at
least one explicitly declared constructor. An instance initializer in an anonymous
class (§15.9.5) can throw any exceptions.

And i tried to create a code as the per the JLS, i not able to suceed.

import java.io.IOException; public class HHH { { // Instance Intializer block throw new IOException(); // line1 } HHH() throws IOException { // explictly declaring the IOException in throws caluse of constructor && i have atleast one default Constructor too } }

i am getting compile time error "Initializer does not complete normally" in line 1, but as per the above quoted JLS Stanza, it should not give compile time error because
1) I have atleast one default constructor
2) that Constructor is also throwing the same Checked, IOException, what i am throwing in intance intializer block

So can someone tell me where i am going wrong here.

HI All,

I came across this question in an interview, how do you make a HashMap as index based like ArrayList. for example the way you insert the elements in the HashMap has to be same as the Way you retrieve elements in the HashMap.

The Answer given by me is "always return the same hashCode so that hashCollision occurs and all the elements will be saved in same bucket." still its not maintaing the index. is there any way to make HashMap index based.

The types of the parameters of the first member method are T1, ..., Tn-1, Tn[].

The types of the parameters of the other method are U1, ..., Uk-1, Uk[].

If the second method is generic then let R1 ... Rp (p ≥ 1) be its type parameters, let Bl be the declared bound of Rl (1 ≤ l ≤ p), let A1 ... Ap be the type arguments inferred (§15.12.2.7) for this invocation under the initial constraints Ti << Ui (1 ≤ i ≤ k-1) and Ti << Uk (k ≤ i ≤ n), and let Si = Ui[R1=A1,...,Rp=Ap] (1 ≤ i ≤ k).

not able to understand what the following equation says, can someone give me the idea.

Ti << Uk (k ≤ i ≤ n), and let Si = Ui[R1=A1,...,Rp=Ap] (1 ≤ i ≤ k).

I am eager to know how you did without final,static,protected keywords

Still there is a draw back in my approach, anyhow i will share how i tried to solve this problem


I have made the class A as abstract and method one() as abstract.

package pck1; abstract class A { abstract public void one(); void two() { } }


package pck2; class B extends A { public void one() { // method one() defined in class A is accessible } // you can't access the method two() as class B is in different package } // so "For class B method one() should be accessible and two() should not be accessible " statement statified
package pck1; // same package as Class A abstract C extends A { public void demo(){ two(); // you can access the method two(); because its in same package; } // here is the drawback of my approach if you override the method one() it will work fine, else if you try to access the method it can't and won't work because its an abstract method } //For class C method one() should not be accessible and two() should be accessible, will be true unless you override method one(); if you override it will be false

Ease up saloni jhanwar

For class B method one() should be accessible and two() should not be accessible
here accessible means i want to access the method one() of class A,So in class B if you override that method(one()) or

access that method(one()) on another method like following
class B extends A{ public void demo() { one(); } } here you will be come to know that method(one()) accessible or not

For class C method one() should not be accessible and two() should be accessible.
above mentioned same is applicable for this statement also, but vise versa.

what craziest thing you wrote above

yeah i used to write craziest thing, because mistakes only makes us to learn perfect things

Thanks for the answers



chain singh

saloni jahnwar wrote:
why you have any doubt?? class B and C should extends class A which is going here this condition satisfied and class C has method two you can check.Does this question state specifically that method two should be of class A ??? i think No.

I think this is your consolidated answer and you are not going to edit it again

ok I will put my question in another way

// you can place it any packages i.e., same Package or difference Package class A { <access specifier> void one() { } <access specifier> void two() { } }



// you can place it any packages i.e., same Package or difference Package class B extends A { @Override void one() { } @Override [b]void two() { // compile time error[/b] } }


// you can place it any packages i.e., same Package or difference Package class C extends <A/B> { @Override [b] void one() { // compile time error [/b] } @Override void two() { } }


Note in class B two() method is giving compilation error and class C one() method is giving compilation error

magnifier rules apply

saloni jhanwar wrote:Use private access modifier while you define two() in class A.It is working.cheers

if you use private access modifier while you define two() in class A, will it satisfy the following statement ?
For class C method one() should not be accessible and two() should be accessible.

BTW I just got the answer that may be one of the way, just wanted to know different ways of thinking and innovative idea's from other people

Hint is I have not used final, static,protected keywords as a access specifiers for the methods. let me post the answer soon.

Hi Saloni,

I Really Appreciate the way you are thinking , Just wanted to confirm few things with you the method two() in class A is package or default specifier that means class on Another package cant override it, here comes an error

package pkg2; import pkg1.a; class B extends A { @Override void two() { // compile time error } } class C extends B { }

let me also try parallel, any new ideas are always welcome.

Regards
S.Sathiya

Hi folks,

I have came across this question in an Interview and not able to answer, The Question is
I have a class A which has one() and two() methods, and I have two another classes B and C both extends A.
The Question is
For class B method one() should be accessible and two() should not be accessible
For class C method one() should not be accessible and two() should be accessible.

use access specifiers, put B and C in another package and so on... But the above condition should meet.

I have tried so many things not able to solve this,

Just wanted to know is it possible ?