+= 和++(预增量)的 Java 运算符优先级

Question

我试图了解以下输出为 5 的 Java 代码的求值顺序:

int a = 1;
a += 2 + ++a;
System.out.println(a);

我对运算符优先级(最高列在前)的理解是:

++ 2
+ 4
+= 14

来自这个列表

http://docs.oracle.com/javase/tutorial/java/nutsandbolts/operators.html

当最终运算符被求值(+=)时，+=的求值开始时a的值不就是2吗？

Answer 1

JLS says the following about the compound assignment operator +=

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.

所以+=用在了

int a = 1;
a += 2 + ++a;

相当于

int a = 1;
a = (int) ((a) + (2 + ++a));

填空，这就变成了

a = 1 + (2 + 2);