Arithmetic promotion

[zarina mohammad]

Question from the new mock exam
http://www.geocities.com/danchisholm2000/june28/exam1.html
Q14

 class Testbyte {
    public static void main (String[] args) {
      byte b=1;
      long l=1000;
      b += l;
      System.out.print(b);
    }
} 

compiles and runs without any error.
i thought they should be an explicit cast to accomodate the long value in byte.

class Testbyte {
    public static void main (String[] args) {
      byte b=1;
      long l=1000;
       b=b+l;
      System.out.print(b);
    }
}  

Now this results in the compiler error(loss of precision).
is b+=l any different from b=b+l?? If No, then why error in the later case?
Is arithmetic promotion taking place only in the former case?
please clear me on this.
thanks
zarina

[Dan Chisholm]

Zarina,
You are correct. If the simple assignment operator had been used as follows, then an explicit cast would be required.

b = (byte)b+1;

The compound assignment operators do not require the explicit cast because the cast is implicit.

E1 op=E2
is equivalent to 
E1=(T)((E1) op (E2))

The rule is stated in section 15.26.2 of the
Java Language Specification.
The question was designed to catch those that know the first rule but not the second.

[Paul Villangca]

Of course the result of the operation would be wildly inaccurate, as precision is lost (long -> byte).

[Dan Chisholm]

Paul,
If the result of the addition falls outside the
range of a byte, then the result would not be accurate. However, if the result falls within the range of a byte, then the result would be exactly correct. Even so, I agree that caution is required when a long value is added to a byte.

[Alan Chong]

b=b+l; // Where can I put (long) ? Nowhere !
Because there are no places for you to put (long), the language designers have to mandate
that the compiler should do the casting for you.
And you are responsible for the lost of precision.