Array chain assignment in Java

Question

Consider the following code snippet in Java. I know that the statement temp[index] = index = 0;in the following code fragment is largely unacceptable, but may be required in some situations and, therefore, you need to know.

package arraypkg;

final public class Main
{
    public static void main(String... args)
    {
        int[]temp=new int[]{4,3,2,1};
        int index = 1;

        temp[index] = index = 0;
        System.out.println("temp[0] = "+temp[0]);
        System.out.println("temp[1] = "+temp[1]);
    }
}

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It displays the following output on the console.

temp[0] = 4
temp[1] = 0

---

I do not understand this statement temp[index] = index = 0;. How does it temp[1]contain 0? How does this appointment happen?

Answer 1

Assignment is performed ( temp[index] = (index = 0)), right associative.

But first, the expression temp[index]is evaluated for the LHS variable. At this time, indexit is still 1. Then RHS ( index = 0) is executed .

Answer 2

. temp [index] = index = 0 AND temp [index]. . 0.

Answer 3

What this line does is say that it temp[index]should equal indexafter indexa value is assigned 0.

This is why this syntax is in most cases unacceptable. It is difficult to read, and most people do not understand this.

Answer 4

you set both temp [1] and the index at '0' it works from left to right. think donkey [index / * * /]