为什么以下代码不会抛出IndexOutOfBoundsException，并打印出9 9 6？

I am new to java. I had a doubt.

我是java的新手。我有一个疑问。

class ArrTest{ 
  public static void main(String args[])
{ 
    int   i = 0; 
    int[] a = {3,6}; 
    a[i] = i = 9; 
    System.out.println(i + " " + a[0] + " " + a[1]); // 9 9 6
  } 
}

2 个解决方案

#1

This is another good example the great Java evaluation rule applies.

这是伟大的Java评估规则适用的另一个好例子。

Java resolves the addresses from left to right. a[i] which is the address of a[0], then i which is the address of i, then assign 9 to i, then assign 9 to a[0].

Java从左到右解析地址。 a [i]是[0]的地址,然后i是i的地址,然后将9分配给i,然后将9分配给a [0]。

IndexOutOfBoundsException will never be throw since a[0] is not out of bound.
The misconception is a[9], which is against the left-to-right-rule

IndexOutOfBoundsException永远不会抛出,因为[0]没有超出范围。误解是[9],这是违反从左到右的规则

#2

It should not.

它不应该。

a[i] = i = 9 (makes i equal to 9) a[0] should also be 9 since you assigned 9 to it (a[i] = i = 9), Initially a[0] was 3 and a[1] is 6 (initial value (int[] a = {3, 6};)

a [i] = i = 9(使i等于9)a [0]也应该是9,因为你为它分配了9(a [i] = i = 9),最初a [0]是3和a [ 1]是6(初始值(int [] a = {3,6};)

You should get 9 9 6.

你应该得到9 9 6。

If you do a[2] then it will give you an IndexOutOfBoundsException.

如果你做[2]那么它会给你一个IndexOutOfBoundsException。

#1