int -> short (OK) ; short -> byte (not ok ) Why?..

Hi,
As 32 bits int can be implicitly narrowed to 16 bits short,why its not possible to have 16 bits implicit conversion to 8 bits byte.
ex:
short s= 24 //OK : 16 bits <- 32 bits
byte b=s; //NOT OK WHY?? : 8 bits <- 16 bits
Prasad Ballari.

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<html><body>
Listen Prasad there is a difference between primitve reference conversion and a constant values conversion. Here is your ex:
<code>
short s= 24 //OK : 16 bits <- 32 bits A conversion from constant
byte b=s; //NOT OK WHY?? : 8 bits <- 16 bits A conversion from primitive refernce or variable
Here is the code that shows an integer type implicitly converted to byte:
public class narrow{
public static void main(String args[]){
byte b=24;
System.out.println(b);
}
}
</code>
see how it works </body></html>

This should also help you...
http://enthuware.com/jqplus/FAJQuestions.html#CONVERSIONS
-Paul.
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Hi Prasad,
if you assign 24 to a byte or short variable, this will work as long as the range of a short or a byte can cover the value. For example an byte b = 300 won't work. Compiler will detect, that he can't put 300 in a byte....

[B]Originally posted by Prasad Ballari:
short s= 24 //OK : 16 bits <- 32 bits byte b=s; //NOT OK WHY?? : 8 bits <- 16 bits
[/B]

When you assign 24 to s, the compiler will know that 24 will never change during the runtime, its an integral literal....
But if you assign s to a byte b, compiler can never be sure, if s will always fit in the byte. That's the reason why he complains. If you change to everything is okay...
hope that helps
as always correct me if i'm wrong
Oliver

Hi guys,
There is no such thing as implicit narrowing conversion.
You cannot implicitly convert an int to a short nor can you convert a short to a byte.
Regarding what Oliver Grass posted :
final int s = 2;
byte b = s;
is OK because s is a CONSTANT and the value can fit into a byte. It is the same as.
byte b = 2;
However you may not use a long , float or double here.
Rgds
Sahir