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)
?
A compound assignment expression of the form
E1 op= E2
is equivalent toE1 = (T)((E1) op (E2))
, whereT
is the type ofE1
, except thatE1
is evaluated only once.[...] the following code is correct:
and results in x having the value 7 because it is equivalent to: