Until today I thought that for example:

`i += j;`

is just a shortcut for:

`i = i + j;`

But what 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)`

?

I’ve tried googling for it but couldn’t find anything relevant.

# 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 to E1 = (T)((E1) op (E2)), where T is the type of E1, except that E1 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.