pls some1 clarify this

[Ishita Gupta]

The following code behaves as i had put in comments.

/** 
case 1: without final on line  5 
Will give compilation error at line 7, 
explicit cast needed for int to byte type
case 2: with final added before int on line 5
compiles, with o/p 100
*/
1 class MyClass
2 {
3  public static void main(String []args)
4  {
5   final int i = 100;
6   byte b = i;
7   System.out.println(b);
8  }
9}

I am unable to undstand why is it that when int i becomes final the code compiles giving the output.
Thanx in advance.
[ Jess added UBB [code] tags to preserve the whitespace, check 'em out! ]
[ January 05, 2003: Message edited by: Jessica Sant ]

[John Lee]

Originally posted by like_java:
The following code behaves as i had put in comments.

/** 
case 1: without final on line  5 
Will give compilation error at line 7, 
explicit cast needed for int to byte type
case 2: with final added before int on line 5
compiles, with o/p 100
*/
1 class MyClass
2 {
3  public static void main(String []args)
4  {
5   final int i = 100;
6   byte b = i;
7   System.out.println(b);
8  }
9}

I am unable to undstand why is it that when int i becomes final the code compiles giving the output.
Thanx in advance.
[ Jess added UBB [code] tags to preserve the whitespace, check 'em out! ]
[ January 05, 2003: Message edited by: Jessica Sant ]

final int i = 100;
byte b = i;
I noticed there is no casting for int i. So this has to do with up casting.

[srini kami]

Hi Like_java
I think it's because when you are assigning int(size32) i to byte(16) i, it needs explicit casting as you are narrowing the conversion.you can make above code work by removing the final as follows with explicit cast.
class MyClass
{
public static void main(String []args)
{
int i = 100;
byte b =(byte) i;
System.out.println(b);
}
}
It is working with final because you can't change value of final(constant), so when you are assigning it to byte, its not assigning it. Thats my best guess

[Jessica Sant]

"like_java"-
two things...
1st thing:
Welcome to the JavaRanch! Please adjust your displayed name to match the JavaRanch Naming Policy.
You can change it here.
2nd thing:

Originally posted by like_java:

case 1: without final on line 5
Will give compilation error at line 7,
explicit cast needed for int to byte type
case 2: with final added before int on line 5
compiles, with o/p 100

final int i = 100;
byte b = i;
when i is marked as final, the compiler is sure of its value -- it knows without a doubt that the value 100 is small enough to be held within a byte -- so it knows there will be no loss of precision, and it'll let you do it.
however, if i is NOT marked as final, the compiler can't be sure of its value, and can't be sure that the value held in i is small enough to go into a byte without losing any precision. So... just to get itself off the hook -- it makes you take the heat if there's a problem (loss of precision) and you have to explicitly cast it.
As Kathy and Bert write in their new certification book:
"A cast is nothing more than your way of saying to the comipler. 'Trust me. I'm a professional. I take full responsibility for anything weird that happens when [you convert that int to a byte].' "
...again welcome to the JavaRanch!
[ January 05, 2003: Message edited by: Jessica Sant ]

[bhatti mansha]

hi like_java
there is another reason
byte have capacity to store 225 and u put in 100
so complier have no problem to complie this code easyly
if there u pass 256 then this is error
and for this u can use costing
thanx
MANSHA BHATTI
i hope you will clear more

[Jessica Sant]

Originally posted by bhatti mansha:
hi like_java
there is another reason
byte have capacity to store 225 and u put in 100
so complier have no problem to complie this code easyly
if there u pass 256 then this is error
and for this u can use costing
thanx
MANSHA BHATTI
i hope you will clear more

Actually a byte can hold -128 to 127. So if you put in a final value between those digits -- no problem. But if you put a final value above 127 or below -128 you'll have to cast it (and you will lose precision -- but because you explicitly cast it -- that's your responsibility).

[jason stark]

I'm wondering if its assigning the same way as
byte b = 100;
as the value of i is final and the compiler is relaxing its usual assignment rules.

[Jessica Sant]

Originally posted by jason stark:
I'm wondering if its assigning the same way as
byte b = 100;
as the value of i is final and the compiler is relaxing its usual assignment rules.

right... because the compiler knows that 100 can fit in a byte without a problem... however if you try the following you'll get an error:
byte b = 128;

[Jose Botella]

This code shows the compiler replacing the access to final primitive variables with the proper value itself, instead of with a reference to the variable.

class Test
{static int i= 1;
static final int j= 2;
static final int k= 0x7FFF;//2^15-1
static final  int m= 0x08000;//2^15
public static void main (String args[]) 
{System.out.println(i);
System.out.println(j);
System.out.println(k);
System.out.println(m);
}
}
/*
Method void main(java.lang.String[])
   0 getstatic #2 Field java.io.PrintStream out
   3 getstatic #3 Field int i
   6 invokevirtual #4 Method void println(int)
   9 getstatic #2 Field java.io.PrintStream out
  12 iconst_2
  13 invokevirtual #4 Method void println(int)
  16 getstatic #2 Field java.io.PrintStream out
  19 sipush 32767
  22 invokevirtual #4 Method void println(int)
  25 getstatic #2 Field java.io.PrintStream out
  28 ldc #5 Integer 32768
  30 invokevirtual #4 Method void println(int)
  33 return
*/

3 getstatic #3 Field int i
accesing i as a variable because it is not final
12 iconst_2
accesing the value of j because it is final
19 sipush 32767
accesing the value of k because is final
28 ldc #5 Integer 32768
accessing the value of m because is final.
This one has to load the value from the constant pool because there are no bytecode able to assign this value directly. "push a short as int" --sipush-- can only assign a signed short (up to 32767 when possitive)