java - 为什么Java的+ =，-=，* =，/=复合赋值运算符不需要强制转换？

Question

直到今天，我还以为例如:

i += j;

只是以下方面的捷径:

i = i + j;

但是，如果我们尝试这样做:

int i = 5;
long j = 8;

然后 i = i + j;将不会编译，但是 i += j;将可以正常编译。

这是否意味着 i += j;实际上是诸如此类的快捷方式 i = (type of i) (i + j)？

Answer 1

与这些问题一样，JLS保留答案。在这种情况下§15.26.2 Compound Assignment Operators。摘录:

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.

§15.26.2引用的一个例子

[...] 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);

换句话说，您的假设是正确的。