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switch case

 
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taken from: http://www.jdiscuss.com
int x = Integer.parseInt(args[0]);
switch(x)
{
case x < 5 : System.out.println("BIG"); break;
case x > 5 : System.out.println("SMALL");
default : System.out.println("CORRECT"); break;
}
this is giving compilation error. but as i know parseInt will convert String into primitive, so it should atlaest compile
 
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Here it won't compile because of case, not because of switch.Case should be compile time constant and should be int here case results in boolean.
 
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In Switch , case statement should be consatant /static final.
 
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Hi

The value given in the case should be constants of type that can implicitly casted to int type.ie, we can use byte, short or int type constants.

In the example, you had used boolean, which can't be casted to int.
 
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The expression used should be a constant one.
//compiles fine
eg: final int x=10;
switch(12):
case x-1:
case x:

If you have just int x=10; it will give compiler error.

But when i tried to compile the below code:
//gives compiler error that const expression is reg
final int x=Integer.parseInt("12");//here i am keeping the value of x as constant.
eg: final int x=Integer.parseInt("12");
switch(12):
case x-1:
case x:

Kindly suggest!
 
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Hello,

Just to add some facts, always remember that switch structure can be used ONLY to check the equality between the switch expression and the case values.
Also remember that only byte, short, int and char primitive types are the ONLY possible expression/case values for a switch.

And one special point, type of the case values must be compatible with the switch expression type.



code given above results in a compile time error. If you look at carefully you can catch the point.




data type of the value 1000 is int and it's not compliant with the switch expression's data type which is byte.

Good Luck!

Regards,
VIRAJ
 
Anoobkumar Padmanabhan
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Hi Vidhya



This problem may arise because x gets value only at runtime. In the definition of switch case construct, it is specified that the case argument should be a compile time constant.

The same error message will get if you use,
int y=20;
final int x=y+1;

But, the error will be resolved if you change that to,
final int y=20;
final int x=y+1;

Because in the latter case, as y is a constant, x also get the value at compile time itself.

I am not sure whether this is cent percentage correct. If i am wrong, please correct it.
 
vidhya suvarna
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Oh. I got the point.
Thanks Anoob.
 
Don't get me started about those stupid light bulbs.
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