Java的+ =， - =，* =，/ =复合赋值运算符 [英] Java's +=, -=, *=, /= compound assignment operators

问题描述

直到今天为止，我还是想：例如：

Until today I thought that for example:

i += j;

只是一个捷径：

i = i + j;

但是，如果我们尝试这个：

But what if we try this:

int i = 5;
long j = 8;

然后 i = i + j; 不编译，但 i + = j; 将编译正确。

Then i = i + j; will not compile but i += j; will compile fine.

这是否意味着事实上 i + = j; 是类似这样东西的快捷方式
i =（i的类型）（i + j） ？

Does it mean that in fact i += j; is a shortcut for something like this i = (type of i) (i + j)?

推荐答案

与这些问题一样，JLS保留了答案。在这种情况下，§15.26.2复合分配操作符< a>。提取：

As always with these questions, the JLS holds the answer. In this case §15.26.2 Compound Assignment Operators. An extract:

以下形式的复合赋值表达式 E1  op =  E2 等效于 E1  = （T）（（E1） （E2）），其中 T 是 E1 的类型，除了 E1 只计算一次。

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.

从§15.26.2

[...]

[...] the following code is correct:

short x = 3;
x += 4.6;

，结果x的值为7，因为它等价于：

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，你的假设是正确的。

In other words, your assumption is correct.

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