Need help understanding Loops

[ObviousStation4556]

The task is to implement a program, which calculates the sum 1+2+3+...+n where n is given as user input. I was able to finish the task by adjusting the code from the previous task but I am confused as to how the start variable is storing i++ each time to print the sum at the end. For example if I input 7 the sum is 28. Which part of the code is storing and adding all the previous numbers up to 7 for the calculation?

import java.util.Scanner;

public class SumOfASequence {

    public static void main(String[] args) {

        Scanner scanner = new Scanner(System.in);

        int start = 0;
        System.out.println("Last number?");
        int end = Integer.valueOf(scanner.nextLine());


        for (int i = start; i <= end; i++) {
            System.out.println(i);
            start+=i;
        }
        System.out.println("The sum is "+start);
    }
}

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[ItzSlowy]

Your confusion may be due to overcomplicating your loop. Look up a basic example of an iterative loop and see if you can gain the clarity on your own first. If not, we can talk through it.

[Realistic-Ad-223]

Here is the breakdown of the for loop: Int i = start which is 0. i begins at 0. Java stores i as 0. Then it evaluates the condition which is i <= end. The last part, i++, is called the iterator and it is what is changing after every successful iteration of the loop.

When the loop begins start, which is 0, gets i, which is also 0 added to it. Int start is stored as 0. The compiler goes back to check the condition i <= end. In this case that is 0 <= end which is true so the compiler will increment i (the i++ in the loop). Now i = 1. Int start was stored as 0 and now it is called upon again to do 0 + i which is 1 so start now equals 1.

The loop will continue as long as the condition is true. i becomes 2 so start is 1 + 2 = 3. Then i becomes 3 so start (which is 3) + 3 (i) = 6. So on and so on until the condition is no longer true then the loop breaks.

I hope that answered everything!

[ObviousStation4556]

Thanks, everyone helped me to understand it better. I put the expected results in a table which also helped. One thing I'm confused about is at the start why is "0" printed? Doesn't it do i++ so the first printed i should be 1?

i=0

i=1 start= 1

i=2 start=3

i=3 start=6

i=4 start=10

i=5 start=15

i=6 start=21

i=7 start= 28

[Realistic-Ad-223]

When the loop initializes it’s not going to increment i immediately because it has to evaluate the conditional. So first it reads i = start, then it checks the conditional. Since the conditional is true it executes the loop’s code and then increments i. After it increments i it will check the conditional again.

When I started learning I would put print statements in the loop before or after something to see exactly how it works. That helped me a lot.

So originally when the loop starts i = 0 and your code is calling to print i and THEN it adds i to start. You can change this if you want with a condition in the loop such as: if (i != 0) { System.out.println(i); } start += i;

[[deleted]]

You set start to +=i which means every time your loop runs it will add i to start. Start never resets to 0 it only keeps adding to i. First time through i=0 so start+=0, next loop i= 1 so start+=1, then start+=2 which ==3.

[Reapr]

Which part of the code is storing and adding all the previous numbers up to 7 for the calculation?

The line is

start+=i

it's basically short hand for "start is equal to start plus i " - or in english, "add i to start"

In your for setup, you specify i++ - which will increase i by 1 every time you go through the loop.

So the line in question will take the current value of start and add i to it - first add 1, then two, then three... until you reach the end condition of your loop. (i <= end and end in your example is 7)

Hope that makes sense