awais syed

you set layout before adding componenet to container.because ech layout has its own policy about setting the component and it is done only when the layout set before adding the componet.

Originally posted by madhur jain:
hi!
why is that if i set borderlayout after adding components to the applet nothing is visible.
this is not the case with other layouts.
madhur.

hi,Younes Essouabni
it is explanation part of your question that
"Is there any equivalent in Java or the only way to achieve it is to use the condition if???"
the second part of your question
"And if there is no way, do you have an idea why are we forced to use an int with a switch? I think that a boolean would have been very helpfull, no? ".
the reason of that i give last time that at compile time compiler create a jump table on basis of case label that is not possible if the boolean expresion is used as a case label.

hi all,
in java switch takes a value that may be byte,short,char and int as argument
and up to my knowledge at compile time the compiler construct a jump table on the basis of case labels and this make the excecuting of switch statement fast at runtime.
and there are situations where use of switch is far effective than if structure.

as per your question you use switch rather then if structure using following code.
we write program using
public class TestSwitch { static void method() { System.out.println("\n\tmethod called"); } static void method10(){ System.out.println("\n\tmethod10 called"); } static void method20(){ System.out.println("\n\tmethod20 called"); } static void method30(){ System.out.println("\n\tmethod30 called"); } public static void main(String [] args) { int i = (int) (Math.random() * 50); System.out.println("\n\tthe value of i is = " + i); System.out.println("\n\tcalling method using if structure"); if( i >=10 && i <= 20) method10(); else if(i >=21 && i <=30) method20(); else if(i>=31 && i <= 40) method30(); else method(); System.out.println("\n\tcalling method using switch"); switch( i/10) { case 1: method10(); break; case 2: switch(i%10){ case 0: method10(); break; default: method20(); break; } break; case 3: switch(i%10) { case 0: method20(); break; default: method30(); break; } break; default: switch(i) { case 40: method30(); break; default: method(); break; } break; } } }
the result od using if structure and switch are same.
[ edited to format code using the [code] and [/code] UBB tags -ds ]
[ June 20, 2002: Message edited by: Dirk Schreckmann ]

in java a class can declare static and non static memebers they can be variables, methods, blocks and classes and also interface(static by default). so any statement that has assignment may come on declaring variables like
class xyz
{
int a = amethod();
public int amethod(){
return 10;
}
}
so in your code if you write it using instance initialization block it will work perfectly
like
[QB]class chkconversion {
class abc {
int k=9;
}
abc a = new abc(),c ;
Object b = a;
{
c = (abc) b; //compile ok
}
}
[ June 09, 2002: Message edited by: awais syed ]
[ June 09, 2002: Message edited by: awais syed ]

because you can not declare method in any method.
but you call this likepublic class StringTest { public static void main(String args[]) { int N = 2; add(n); } public void int add(int N) { return 0x100 + N++ ; } }
[ June 08, 2002: Message edited by: Dirk Schreckmann ]

Originally posted by Christoffer Jeffson:
Stumbled over this at John Hunt's mock exam..
public final class kokos { class Inner { void test() { if (Test4.this.flag) { sample(); } } } private boolean flag = false; public void sample() { System.out.println("Sample"); } public Test4() { (new Inner()).test(); } public static void main(String args []) { new Test4(); } }
What bothers me is the Test4.this.flag reference. How is it possible to reference "Test4" (class) and then "this" (instance)?

is top level class name Test4 either "kokas"

Originally posted by Asif Masood:
I am bit confused from the different behaviour for almost same expresion.

char ch2 = '2';
char ch3 = '1';
ch3 += ch2; //this works fine

But the following expression gives compiler error.
ch3 = ch3 + ch2;
Why it's so.

because java compiler reads
ch3 += ch2 as ch3 = (char)ch3 + ch2;