+= int and double conversion

Question

Possible Duplicate:
Varying behavior for possible loss of precision

int testing = 0;
testing += 2.0

the above code compiles.

where as

int testing = 0;
testing = testing + 2.0;

this code doesn't compile. Any idea why?

Answer 1

Compound assignment has a hidden cast in Java.

JLS 15.26.2 Compound Assignment Operators:

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);

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