Why does this compile

[pankaj patil]

code 1)
final int i=50;
byte b=i;
code 2)
int i=50;
byte b=(byte)i;


in code 1) i dont have to typecast
in code 2) i have to typecast

can any one tell me the reason

with regards
pankaj

[Svend Rost]

Guessing here, so please correct me if im wrong.

in 1) i is final, ie. the compiler knows it'll stay at 50
which isn't the case in 2).

/Svend Rost

[pankaj patil]

but the size of int is more then the byte.

and final dose not allow the value to be changed . is that the final variable allocates only that much size, which is required at the initialization.

[Jesper de Jong]

But the value of the final int i is 50 and cannot be changed. The compiler is smart enough to understand that the fixed value 50 fits into a byte (because the range for byte is -128 to +127). Try this:

final int i = 130;
byte b = i;

This will give you an error.

[Layne Lund]

Originally posted by pankaj patil:
but the size of int is more then the byte.

and final dose not allow the value to be changed . is that the final variable allocates only that much size, which is required at the initialization.

Because the variable is declared as final, the compiler will not allocate any memory for it. As mentioned above, the compiler is also smart enough to see that 50 is small enough to fit in a byte.

Layne