Why does Java perform implicit type conversion from double to integer when using the "plus equals" operator? [duplicate]

Question

Possible Duplicate:
Varying behavior for possible loss of precision

Code Sample A

 public class Test {                                                         
     public static void main(String[] args) {
         int i = 0;
         i = i + 1.5;
     }
 }

Code Sample B

 public class Test {                                                         
     public static void main(String[] args) {
         int i = 0;
         i += 1.5;
     }
 }

Unsurprisingly, compiling A produces the error below. Surprisingly, compiling B produces no error and it appears to behave as if I inserted an explicit cast to integer before the double value 1.5. Why in the world does this happen? This goes against everything I thought I knew!

Test.java:6: possible

 loss of precision

    found   : double
    required: int
            i = i + 1.5;
                  ^
    1 error

Answer 1

It is working as designed. The compound operators add an implicit cast to the operation. Otherwise you have to use an explicit cast.

More info?

http://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.26.2

Answer 2

According to the Java language specification, section 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. For example, the following code is correct:

short x = 3;
x += 4.6;

and results in x having the value 7 because it is equivalent to:

short x = 3;
x = (short)(x + 4.6);