Java Question

[Ajay Eyunni]

Please look into the following program:
public class Test8
{
public static void main(String arg[])
{
final byte b = 1;
char c = 2;
short s = 3;
int i = 4;

c = b;
s = b;
i = b;
s = c * b;
}
}

How character can accept byte by declaring byte final?

[Keith Lynn]

Because by making it a final and having an initializer when its declared makes it a compile-time constant.

Since the value is known at compile-time and is a valid char, it can be assigned to a char.

[Deepak Bala]

s = c * b; would create an error though. Possible loss of precision since the result of the multiplication is an int and it cannot be implicitly converted into a short.