The difference between + + and + + operators

We know:

int i = 5;

Long j = 7;

i = i + j cannot compile, but i + = j can compile and run, result i = 12.

This is because:

i + = J equates to i = (int) (I+J);

The conclusion is that for compound assignment expressions, E1 op= E2 (such as i + = j; I-= j, etc.), is actually equivalent to E1 = (t) ((E1) op (E2)), where T is the type of the element E1.

This question, in fact, has been answered in the official documentation. Official document Address §15.26.2 Compound assignment Operators

The difference between + + and + + operators