byte question

byte b = 0;
b += 1; //1 no error
b = b + 1; //2 error

why //1 has no error?

Every arithmetic operation evaluates atleast to an integer ( primitive int ). You have to do a explicit type cast in second line, but += does this implicitly.

HTH

oh, thanks Srinivasa!

Don't forget, there are also rules about automatic implcit narrowing promotion.

Fore example.

byte b = 3; //Ok,

int i = 3;
byte c = i; //Not ok, not resolved at compile time.

So for implicit automatic promotion the following apply:

a) The operand to the right of the assignment operator must be a byte, char, short, or int.
b) Operand on the right must be in the range of the operator on the left
c) Operator on the left must be a byte, short, or char.

Anything else required an implicit cast.

int x = 3L; //Compile error, need a cast

int x = (int) 3L; //OK.

Hi,
int i=3;
byte c = i;

a) The operand to the right of the assignment operator must be a byte, char, short, or int.
b) Operand on the right must be in the range of the operator on the left c) Operator on the left must be a byte, short, or char.

Anything else required an implicit cast.

The above example...satisfies all three conditions...
Then why the compilation error..

You have missed the keyword "final"

put int i as

final int i=3..and then try..

Originally posted by A Kumar:
Hi,
int i=3;
byte c = i;

The above example...satisfies all three conditions...
Then why the compilation error..

You have missed the keyword "final"

put int i as

final int i=3..and then try..

Originally posted by Parthiban Murugesan:
[QB][/QB]

Hi the solution adding "final" may not work if

final int b =130 :-) as this would ask for type casting.

Hi,

Hi the solution adding "final" may not work if final int b =130 :-) as this would ask for type casting.

b) Operand on the right must be in the range of the operator on the left

It is not in the range so...