byte i = 0;
i = i + 1; // 编译错误，因为自动转换int问题而出错

byte i = 0;
i += 1; // 正常编译，可以理解这是将i加上一个单位（i类型的单位）

15.26.2 
Compound Assignment Operators
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);