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 thisi = (type of i) (i + j)
?
Answer
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 toE1 = (T)((E1) op (E2))
, whereT
is the type ofE1
, except thatE1
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.
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