Adding two byte in by using Different Syntax

[Prakash Rai]

class Test{
	public static void main(String[] args) {
		 byte b=10;
		 byte b1=20;
		//1.   b+=b1; Can I write b=b+20; ? It is not giving any error
		//2.  b=b+b1; This statement is giving error but why not above one is giving error
		}
}

Please Help me ::

Q.Why 1st one does not give any error and 2nd is giving error ?
Thanks in Advance:
Note:
1.also can be written b+=b1 or b=b+b1;

[Henry Wong]

Prakash Rai wrote:

class Test{
	public static void main(String[] args) {
		 byte b=10;
		 byte b1=20;
		//1.   b+=b1; Can I write b=b+20; ? It is not giving any error
		//2.  b=b+b1; This statement is giving error but why not above one is giving error
		}
}

Please Help me ::

Q.Why 1st one does not give any error and 2nd is giving error ?
Thanks in Advance:
Note:
1.also can be written b+=b1 or b=b+b1;

The Java language specification has an implicit cast in the expanding expression. See section 15.26.2 ....


http://java.sun.com/docs/books/jls/third_edition/html/j3TOC.html


Henry

[Prakash Rai]

Sorry Henry But I did't my ans? please tell me the point...

Here In one case it is doing Implicit casting but why not it is doing in 2nd case.?

[Jeff Verdegan]

Prakash Rai wrote:Sorry Henry But I did't my ans? please tell me the point...

Here In one case it is doing Implicit casting but why not it is doing in 2nd case.?

He did give you the answer: It's because the JLS says so.

http://java.sun.com/docs/books/jls/third_edition/html/expressions.html#15.26.2:

A compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T)((E1) op (E2)), where T is the type of E1, except that E1 is evaluated only once.

http://java.sun.com/docs/books/jls/third_edition/html/expressions.html#15.18.2:

Binary numeric promotion is performed on the operands (§5.6.2).

[Prakash Rai]

Jeff Verdegan wrote:

Prakash Rai wrote:Sorry Henry But I did't my ans? please tell me the point...

Here In one case it is doing Implicit casting but why not it is doing in 2nd case.?

He did give you the answer: It's because the JLS says so.

What IS JLS.. Please tell me some Appropriate ans...

[Matthew Brown]

The JLS is the Java Language Specification.

It says (in the section that Henry referenced) that:

b += b1;

is equivalent to

b = (byte)(b + b1);

It's this cast (implicit, because it's added automatically) that makes the difference. Any arithmetic operation between two integer types that are smaller than int results in an int. So a cast is needed to be able to assign it back to a byte.

[Prakash Rai]

Jeff Verdegan wrote:

Prakash Rai wrote:Sorry Henry But I did't my ans? please tell me the point...

Here In one case it is doing Implicit casting but why not it is doing in 2nd case.?

He did give you the answer: It's because the JLS says so.

http://java.sun.com/docs/books/jls/third_edition/html/expressions.html#15.26.2:

A compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T)((E1) op (E2)), where T is the type of E1, except that E1 is evaluated only once.

http://java.sun.com/docs/books/jls/third_edition/html/expressions.html#15.18.2:

Binary numeric promotion is performed on the operands (§5.6.2).

Thanks A lot Jeff Verdegan...

[Jeff Verdegan]

Prakash Rai wrote:

Jeff Verdegan wrote:

Prakash Rai wrote:Sorry Henry But I did't my ans? please tell me the point...

Here In one case it is doing Implicit casting but why not it is doing in 2nd case.?

He did give you the answer: It's because the JLS says so.

What IS JLS.. Please tell me some Appropriate ans...

Please take the time to carefully read the answers that people give you, and to follow links that they provide. Henry said:

The Java language specification has an implicit cast in the expanding expression. See section 15.26.2 ....

http://java.sun.com/docs/books/jls/third_edition/html/j3TOC.html