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I don't understand why this isn't working. Everything looks right, and when I say everything out loud it makes sense, but something is wrong with it and the wizard has nothing to say about it. Can an actual person tell me what's wrong with it?

package en.codegym.task.pro.task04.task0403; import java.util.Scanner; /* Summation */ public class Solution { public static void main(String[] args) { //write your code here Scanner console = new Scanner(System.in); String inPut = console.nextLine(); int sum = 0; while (inPut != "ENTER") { String num = console.nextLine(); int x = Integer.parseInt(inPut); sum = (sum + x); } System.out.println("sum"); } //System.out.println(x); }

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Qin-1999

Level 22

19 December 2023, 06:25

Try this one. Good luck

public static void main(String[] args) {
    Scanner console = new Scanner(System.in);
    int sum = 0;
    while (true) {
        String inPut = console.nextLine();
        if (!"ENTER".equals(inPut)) {
            Pattern p = Pattern.compile("\\d+");
            Matcher m = p.matcher(inPut);

            while (m.find()) {
                int num = Integer.parseInt(m.group());
                sum += num;
            }
        }
        if ("ENTER".equals(inPut)) {
            System.out.println(sum);
            return;
        }
    }
}

Thomas

Level 41 , Bayreuth, Germany

17 December 2023, 21:09

You start off pretty cool, but then things drift apart... first the two little problems 1. If you compare objects you usually are not interested if the share the same memory address inside the heap (done using ==). More interesting is if they are equal (o1.equals(o2)). And after your last questions I thought you're the master of equality ;) 2. At the end you do not print the value of the variable sum but instead the word 'sum' 3. The show stopper: In line 13 you ask for an inPut (good) Line 16: check if the inPut does not equal 'ENTER' (good - if the little problem gets fixed) Line 18: you ask for another input and assign that to second variable num (wrong - see below) Line 19, 20: parsing of the original input and adding to sum (good) now here is missing something... the inPut variable hasn't changed inside the loop and if you start a new iteration the loop head checks the original value of inPut. So it either never enters the loop body or runs forever. (hint: move line 16 to a better place and change the variable you assign the value to) spoiler alarm: don't scroll further before you've fixed your code ;) the same as you have written just a little bit more compact (using a for loop you could make it a oneliner)

int sum = 0;

String input;
while (!"ENTER".equals(input = console.nextLine())) {
    sum += Integer.parseInt(input);
}

// for (String input; !"ENTER".equals(input = console.nextLine()); sum += Integer.parseInt(input));

Matthew Riggs

Level 6 , United States of America, United States

18 December 2023, 19:26

Why does it need (sum += Integer) instead of (sum = Integer + sum)? And (!ENTER.equals) vs (!=)? Are they just differences between Java and Python? (I came from Python) Or am I not not fully understanding something?

Thomas

Level 41 , Bayreuth, Germany

18 December 2023, 20:21

That's the regular form to write a sum sum = sum +1; you can shorten this to sum += 1; if you just raise by one you also can use post- or prefix increment sum++ or ++sum You just use what you're comfortable with. The rest however you should be able to understand at least. equals vs. == When you create an object it is saved somewhere inside the memorey (on the heap). The constructor returns a reference, like:

Rectangle rect = new Rectangle(0, 0, 10, 20);

rect is the reference and knows where on the heap the object is. rect knows the capabilities of the object and you can use it to handle the object but it is not the object. If two references point to the same object, then they are identical. To check if two references are the same, then you use == like:

Rectangle rectCopy = rect;
System.out.println(rect == rectCopy);

To verify that I talk the truth we create a second, identical object (rectangle with the same dimensions) and then compare references. They should be different (false)

Rectangle rect2 = new Rectangle(0, 0, 10, 20);
System.out.println(rect == rect2); // false

So rect is pointing to the first object and rect2 two the second object we created. Both have different addresses on the heap and if you compare these references the are not the same. If you now want to check if the objects are equal (what the coder of the class of that object considers to be equal) then you use the equals method. Here it would compare x, y, width and height. If these properties are the same then the two rectangles are considered to be the same. This functinality is not magically there. You need to code that for each class.

System.out.println(rect.equals(rect2)); //true

Thomas

Level 41 , Bayreuth, Germany

18 December 2023, 20:29

The problem to understand this is using the String class to explain this. String uses a technique that is called interning and is nothing different than caching. If you eg. create to Strings...

String a = "Hello";
String b = "Hello";
System.out.println(a == b); // true

then the JVM puts the first Hello into the cache. When you try to add the second Hello, then the JVM recognizes there is already a string with this value, marks the second for garbage collection and returns the address of the first. Means this two strings have the same address in memory and comparing using == will work. However this not always is true as you can create Strings using the constructor and that is bypassing String interning (hence no caching). These strings don't share the same address and == fails

String c = "CG";
String d = new String("CG");
System.out.println(c == d); // false

To sum it up, == is for comparing memory addresses and equals compares object values