about “+=” operator in Java

最新推荐文章于 2023-12-26 00:27:55 发布

原创最新推荐文章于 2023-12-26 00:27:55 发布 · 622 阅读

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本文探讨了Java中不同类型的变量进行加法运算时的行为差异，特别是整型变量与长整型变量之间的混合运算。通过具体的例子说明了+=运算符的工作原理，并引用了《Java语言规范》中的相关规定。

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I found this interesting question on stackoverflow:

Many of us may believe that:

<span style="white-space:pre">	</span>i +=j;

is just a shortcut for

<span style="white-space:pre">	</span>i=i+j;

But let's try this:

<span style="white-space:pre">	</span>int i=5;
<span style="white-space:pre">	</span>long j=8;

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

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

Answer:

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

An extract:

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.

An example cited from §15.26.2

[...] the following code is correct:

<span style="white-space:pre">	</span>short x = 3;
<span style="white-space:pre">	</span>x += 4.6;

and results in x having the value 7 because it is equivalent to:

<span style="white-space:pre">	</span>short x = 3;
<span style="white-space:pre">	</span>x = (short)(x + 4.6);

In other words, the assumption is correct.