casting doubt

[lusha tak]

Hello all!
byte b=16;
//ok as the source value is well with in the range of byte
now consider the case
Byte byteobj=new Byte((byte)16);
why do we need to go for type casting in the above line???
Please explain
Regards
Lusha

[George Pincharoen]

16,by default is int
and int is not an argument of Byte constructors (it takes byte and String) so,you need to cast 16 to byte before
create new instance of Byte

[lusha tak]

nut why don't we require a cast in th eline one.
i.e.
byte b=16//works ok even without cast

[swati bannore]

It works finr if u send variable b to the constructor.
i.e. Byte obj = new Byte (b);
But when u write 16, it's a integer literal by default.So the cast is required as mentioned in the second post

[lusha tak]

Thanks all!
Lusha

[Thomas Paul]

byte b=16; works because even though 16 is an int, the compiler knows that 16 can fit in a byte. This doesn't work:
int a=16;
byte b=a;

[jayu jadhav]

Then look at the code:
int i= -20;
final int j=20;
final int k=i;
byte b1=j; //works
byte b2=(byte)i; //cast required
byte b3=(byte)k; //cast needed
Why is cast not required when source is final int? i. e. can you explain about b1 and b3?
Thanks.
[This message has been edited by jayu jadhav (edited July 18, 2001).]

[Eric Pramono]

Hi Jayu,
I don't think that your statement is correct.
Consider the following code:

public class X {
    public static void main (String[] args) {
        final int a=41;
        final int b=129;
        int c=11;
        byte d=(byte)b;
        byte e=(byte)c;
    }
}

the b variable will need to be cast explicitly into byte, otherwise it'll give a compiler error.
After running javap -c on the class, here's what I got:
0 bipush 41
2 istore_1
3 sipush 129
6 istore_2
7 bipush 11
9 istore_3
10 bipush -127
12 istore 4
14 iload_3
15 i2b
16 istore 5
18 return
please note the difference in line 0 and line 3, where:
final int a=41
will be stored using bipush (assumably byte-push), while:
final int b=129
will be stored using sipush (assumably short-push)
another interesting thing to note is when you explicitly cast a final variable then it'll do the push command, but if you explicitly cast a non-final variable then it'll try to load & convert it..
it's much like assembler where you have macro and compiler simply just replace all instance of "final variable" with the value assigned.
correct me if i'm wrong.
- eric

[Thomas Paul]

If the final variable is within a range that does not cause loss of precision then the compiler will allow it. The reason is that the compiler can treat the final variable as a constant since it knows that it will never be changed.
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Tom - SCJP --- Co-Moderator of the Programmer Certification Forums