Java Puzzler - casting a double to int

Question

int anInt = 1;
double aDouble = 2.5;

anInt = anInt + aDouble; // Error - need to cast double to int

anInt += aDouble; // This is ok. Why?

anInt = aDouble; // This is also an error.

anInt = 1 + aDouble; // This is also an error.

So my questions is: Why is it not a compile error to do anInt += aDouble?

@Gunaseelan i guess he is asking why the `int = int + double` does throw an error while `int += double` doesn´t — SomeJavaGuy, Mar 03 '15 at 13:59
possible duplicate of [Java += operator](http://stackoverflow.com/questions/8710619/java-operator) — Florent Bayle, Mar 03 '15 at 14:02
Hate interview questions: their relation to practice is zero to none. — Denis Kulagin, Mar 03 '15 at 14:11

score 8 · Accepted Answer · answered Mar 03 '15 at 14:02

Three of the four cases properly report an error. Compound assignment is the only exception from the rule. Java Language Specification, part 15.26.2, explains why:

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);

As you can see, the error is avoided by implicit insertion of a cast.

Yes. I came across it in my code and was just surprised the compound assignment didn't need a cast. Thanks for the answer. — Tony Narvarte, Mar 04 '15 at 16:09

Java Puzzler - casting a double to int

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