switch case and wrapper objects

[Sreedevi Vinod]

hi

I was trying out this code

public static void main(String args[]) 
{ 
int t=Integer.parseInt(args[0]);
final int i=10;
final Integer j=12;
switch(t){
case i:
System.out.println("got int");
break;
case j:
System.out.println("got Integer");
break;
}
}

Why is j not acceptable as a valid case, since it is final (just as i is)?

Thanks
Devi

[Swapna James]

Hi Devi,

The only type that a switch can evaluate is the primitive int. The values that can be promoted to an int, that is byte, short, char are also acceptable. Here you are trying to match an Integer Wrapper object to the switch statement. That is why you are getting that error.

I hope this explains your doubt. And you rest of you guys.... correct me if I am wrong.

Swapna.

[Stephen O'Kane]

Sounds correct to me Swapna.

The only other thing that Devi could do is to type

case j.intValue();

in the switch statement to get a primitive int.

Sok

[A Kumar]

Hi

Sorry Stephen..

it gives a compiler error...

case expressions must be constant expressions...

what does that mean???

Regards

[Swapna James]

Hi Guys,

An excerpt from K&B book chapter 4:

The only argument a case can evaluate is one of the same type as switch can use,with one additional�and big�constraint: the case argument must be final! The case argument has to be resolved at compile time, so that means you can use only a literal or final variable.

Hope this clears your doubt.

Swapna.

[sangeetha balasubramaniam]

Hi,

case argument can only take a literal or final variable.

eg 1:
byte a = 20;
switch(a){
case 20:
{
System.out.println("20");
break;
}
case 127:
{
System.out.println("127");
break;
}
}


eg 2:
final int one = 1;
final int two = 2;
int a = 2;
switch(a){
case one:
{
System.out.println("one");
break;
}
case two:
{
System.out.println("two");
break;
}
}eg 1:

[Sreedevi Vinod]

That is not right. The switch statement does allow wrappers of the numeric types, since autoboxing is now supported. Then why doesn't the case statement take a wrapper type as argument when it is declared final and hence is a constant ?

Thanks
Devi

[Joshua Smith]

Also, remember that Java 5 adds enums as valid arguments to case statements.

Details here:
http://java.sun.com/j2se/1.5.0/docs/guide/language/enums.html

Josh

[Sreedevi Vinod]

What I mean is that the below code allows t to be an Integer, but will not compile if I make j an Integer ( though j is final). Any ideas ?

public static void main(String args[]) 
{ 
Integer t=Integer.parseInt(args[0]);
final int i=10;
final int j=12;
switch(t){
case i:
System.out.println("got int");
break;
case j:
System.out.println("got Integer");
break;
}
}
 

Thanks
Devi
[ September 01, 2005: Message edited by: Sreedevi Vinod ]

[Steve Morrow]

What I mean is that the below code allows t to be an Integer, but will not compile if I make j an Integer ( though j is final). Any ideas?

The "case values" must be constant expressions, which by definition do not include values of Integer types.