Int can be incremented by a double value

Question

This code seems to work in Java, violating everything I thought I knew about the language:

int x = 0;
x += 7.4;

x now has the value 7. Of course, one can't just write int x = 7.4, so this behavior seems strange and inconsistent to me.

Why did the developers of Java choose such a behavior?

The question that mine was marked as a duplicate of was actually answering the "what happens" part, but not my main question: what the rationale is.

Answer 1

The operators for numbers do all kinds of casting which in this case converts the 7.4 double to a 7 int by rounding it.

What you have here is a Compound Assignment Operators

So what really gets executed is

x= (int)(x + 7.4)

Since x is an int and 7.4 x gets converted to double vs a Binary Numeric Promotion so you get 7.4 as an intermediate result.

The result (a double) is then cast and therefore subject to a Narrowing Primitive Conversion which rounds it to 7

Regarding the new question: Why was it done this way?

Well you can argue long if implicit conversions are a good or bad thing. Java went some kind of middle road with some conversions between primitives, their boxed types and Strings.

The += operator then has a rather simple and straight forward semantics. It really only looks strange if you consider it an increment by operator, instead of what it really is: a shorthand for a combination of operator and assignment.

Answer 2

Sometime back only i read about it It will be actually X= (int)(x + 7.4)