Until today I thought that for example:
i += j;
is just a shortcut for:
i = i + j;
But what if we try this:
int i = 5; long j = 8;
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)?
I’ve tried googling for it but couldn’t find anything relevant.
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.