3gZ2jA
diff --git a/Collapse file
‎contents/java-operator.md‎
Copy file name to clipboardExpand all lines: contents/java-operator.md
+5-5Lines changed: 5 additions & 5 deletions
Display the source diff
Display the rich diff b/Collapse file
‎contents/java-operator.md‎
Copy file name to clipboardExpand all lines: contents/java-operator.md
+5-5Lines changed: 5 additions & 5 deletions
Display the source diff
Display the rich diff
@@ -8,17 +8,17 @@ i += j 等同于 i = i + j;
 int i = 5;
 long j = 8;
 ```
-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
-这是否意味着，i += j,实际是等同于 i= (type of i) (i + j)呢？
+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
+这是否意味着，i += j，实际是等同于 i= (type of i) (i + j)呢？
 
 ###回答
-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 )
+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/se8/html/jls-15.html#jls-15.26.2)
 
 
 再照搬下官方文档的说明
 
 
-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。
+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。
 
 举例来说，如下的代码
 ```java
@@ -33,4 +33,4 @@ x = (short)(x + 4.6);
 
 
 stackoverflow链接
-http://stackoverflow.com/questions/8710619/java-operator
+http://stackoverflow.com/questions/8710619/java-operator