Why don't Java's +=, -=, *=, /= compound assignment operators require casting?
assignment-operator, casting, java, operators, variable-assignment
Solution
As always with these questions, the JLS holds the answer. In this case §15.26.2 Compound Assignment Operators. An extract:
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.
An example cited from §15.26.2
[...] 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);
In other words, your assumption is correct.
Problem
Until today, I thought that for example: ``` i += j; ``` Was just a shortcut for: ``` i = i + j; ``` But if we try this: ``` int i = 5; long j = 8; ``` Then `i = i + j;` will not compile but `i += j;` will compile fine. Does it mean that in fact `i += j;` is a shortcut for something like this `i = (type of i) (i + j)`?