开发者

Why the following code does not throw IndexOutOfBoundsException, and print out 9 9 6?

开发者 https://www.devze.com 2023-03-23 18:58 出处:网络
I am new to java. I had a doubt. class ArrTest{ public static void main(String args[]) { inti = 0; 开发者_如何转开发int[] a = {3,6};

I am new to java. I had a doubt.

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


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

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].

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


It should not.

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};)

You should get 9 9 6.

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

0

精彩评论

暂无评论...
验证码 换一张
取 消