Logic of for loop using with array to access elements of the array in java

Question

My code to access elements of array using for loop. The output of the program is [19,17,15] which are the elements of array int a[] = { 12, 15, 16, 17, 19, 23 }. Output after following code is written:

public class Tester {
    public static void main(String s[]) {
        int a[] = { 12, 15, 16, 17, 19, 23 };
        for (int i = a.length - 1; i > 0; i--) {
            if (i % 3 != 0) {
                --i;
            }
            System.out.println(a[i]);
        }
    }
}

Algorithm:

Iteration 1: i=5 so a[5]=23. "if statement" gets true and --i execute. so i=4. Hence, a[4]=19 will get print as first element in output.
Iteration 2: i=3 so a[3]=17. "if statement" gets true again and --i should execute but it skips that condition as I tried using debugging tool to see how it is working. And output is a[3]=17 which is wrong I think.

Could you help me in understanding why it gives 17 as output? As in my opinion it should skip that part.

If i = 3, then i % 3 is 0. That would mean your if statement becomes false and not true. — DanielBarbarian, Commented May 13, 2022 at 9:43
And since the if statement is false, that would explain the behavior you are seeing. — DanielBarbarian, Commented May 13, 2022 at 9:44
Your if statement doesn't check a[i] % 3 != 0 - it checks i % 3 != 0. In the first iteration i is 5 and i % 3 != 0 is true. In the second iteration i is 3 and i % 3 != 0 is false and therefore i is not decremented and a[3] gets printed. — Thomas Kläger, Commented May 13, 2022 at 9:45
a%b is the modulo operator. It returns the reminder of a/b — xImperiak, Commented May 13, 2022 at 9:48
Your problem is that you decrement i twice (once as a result of the condition and again as an an unconditional part of your forloop). Instead replace the if body with continue. — Sherif, Commented May 13, 2022 at 9:53

WJS · Accepted Answer · 2022-05-13 12:02:06Z

0

Here is a step by step explanation. (i % 3 != 0) checks to see if i is not divisible by 3. Also note that in this context, your post and pre-decrements of i are not relevant as the outcome would be the same no matter how they are decremented.

 i = 5;
 i not divisible by 3 so i-- = 4
 print a[4] = 19
 i-- at end of loop = 3
 i is divisible by 3 so i remains 3
 print a[3] = 17
 i-- at end of loop = 2
 i not divisible by 3 so i-- = 1
 print a[1] = 15
 i-- = 0 so loop terminates.

edited May 13, 2022 at 12:02

answered May 13, 2022 at 11:49

WJS

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Boaz Lev · Accepted Answer · 2022-05-13 09:52:10Z

-1

In iteration 2, where i=3 the condition is false, since 3 % 3 = 0

answered May 13, 2022 at 9:52

Boaz Lev

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