{Tips For Java} Compound assignment operator.

Question

i += j;
is not 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.

Because: (JLS)
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.

For example, 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);

Answer 1

You should be appreciated for posting concepts like this. Keep doing this. I already am aware of this concept but many are not and this will help them.

Answer 2

Now that's something I didn't know. Thanks for the information

Answer 3

thank you