Why don't Java's +=, -=, *=, /= compound assignment operators require casting?

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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)?

Answer

As always with these questions, the JLS holds
the answer. In this case href="http://docs.oracle.com/javase/specs/jls/se8/html/jls-15.html#jls-15.26.2">§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 href="http://docs.oracle.com/javase/specs/jls/se8/html/jls-15.html#jls-15.26.2">§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.