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)
?
Best Answer
As always with these questions, the JLS holds the answer. In this case §15.26.2 Compound Assignment Operators. An extract:
An example cited from §15.26.2
In other words, your assumption is correct.