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Commit 98f7c44

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Java += 操作符实质
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‎README.md‎

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> 基础语法
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* [Java是按值传递还是按引用传递](https://github.com/giantray/stackoverflow-java-top-qa/blob/master/contents/is-java-pass-by-reference-or-pass-by-value.md)
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* [Java += 操作符实质](https://github.com/giantray/stackoverflow-java-top-qa/blob/master/contents/java-operator.md)
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> 代码规范
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### 待翻译问题链接(还剩0问题)
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- [Why is processing a sorted array faster than an unsorted array?](http://stackoverflow.com/questions/11227809/why-is-processing-a-sorted-array-faster-than-an-unsorted-array)
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- [Why is subtracting these two times (in 1927) giving a strange result?](http://stackoverflow.com/questions/6841333/why-is-subtracting-these-two-times-in-1927-giving-a-strange-result)
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- [Is Java "pass-by-reference" or "pass-by-value"?](http://stackoverflow.com/questions/40480/is-java-pass-by-reference-or-pass-by-value)
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- [Java += operator](http://stackoverflow.com/questions/8710619/java-operator)
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- [Avoiding "!= null" statements in Java?](http://stackoverflow.com/questions/271526/avoiding-null-statements-in-java)
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- [Proper use cases for Android UserManager.isUserAGoat()?](http://stackoverflow.com/questions/13375357/proper-use-cases-for-android-usermanager-isuseragoat)
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- [Differences between HashMap and Hashtable?](http://stackoverflow.com/questions/40471/differences-between-hashmap-and-hashtable)

‎contents/java-operator.md‎

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##Java += 操作符实质
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###问题
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直到今天,我都一直以为:
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i += j 等同于 i = i + j;
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但假设有:
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int i = 5;
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long j = 8;
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这时 i = i + j不能编译,但i += j却可以编译。这说明两者还是有差别的
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这是否意味着,i += j,实际是等同于 i= (type of i) (i + j)呢?
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###回答
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这个问题,其实官方文档中已经解答了。 请看这里。[§15.26.2 Compound Assignment Operators](http://docs.oracle.com/javase/specs/jls/se5.0/html/expressions.html#15.26.2 )
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再照搬下官方文档的说明
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对复合赋值表达式来说,E1 op= E2(诸如i += j;i-=j等等),其实是等同于E1 = (T)((E1) op (E2)),其中,T是E1这个元素的类型。
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举例来说,如下的代码
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short x = 3;
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x += 4.6;
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等于
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short x = 3;
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x = (short)(x + 4.6);
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stackoverflow链接
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http://stackoverflow.com/questions/8710619/java-operator

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