Issue with the Loop | Together Java | Page 1

tender nestBOT Nov 22, 2023, 9:21 PM

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<@&987246399047479336> please have a look, thanks.

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Without seeing your code, it's difficult to pinpoint the exact issue. However, here's an example of how you can use two variables (i and tmp) to add numbers in a loop:

public class Main {
    public static void main(String[] args) {
        int i = 0;
        int tmp = 0;
        
        while (i < 10) {
            tmp += i;
            i++;
        }
        
        System.out.println("Sum: " + tmp);
    }
}

In this example, the loop starts with i as 0 and tmp as 0. It continues until i reaches 10. Inside the loop, the value of i is added to tmp, and then i is incremented by 1. Finally, outside the loop, the sum stored in tmp is printed.

If your code is similar to this and still getting stuck, please provide your code so that we can help you better.

pale sentinel Nov 22, 2023, 9:24 PM

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don't use nextInt()

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instead do nextLine()

silver gorge Nov 22, 2023, 9:24 PM

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Doesnt next line read String

pale sentinel Nov 22, 2023, 9:26 PM

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yeah but nextInt does not consume the next line so you run into issues like you're seeing

silver gorge Nov 22, 2023, 9:26 PM

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This way it worked

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But I cant display count of number now

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pale sentinel Nov 22, 2023, 9:27 PM

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Integer.parseInt(scanner.nextLine()); just do it this way

tender nestBOT Nov 22, 2023, 9:58 PM

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@silver gorge

Mixing any nextXXX method with nextLine from the Scanner class for user input, will not ask you for input again but instead result in an empty line read by nextLine.

To prevent this, when reading user input, always only use nextLine. If you need an int, do

int value = Integer.parseInt(scanner.nextLine());

instead of using nextInt.

Assume the following:

Scanner scanner = new Scanner(System.in);

System.out.println("Enter your age:");
int age = scanner.nextInt();
System.out.println("Enter your name:");
String name = scanner.nextLine();

System.out.println("Hello " + name + ", you are " + age + " years old");

When executing this code, you will be asked to enter an age, suppose you enter 20.
However, the code will not ask you to actually input a name and the output will be:

Hello , you are 20 years old.

The reason why is that when you hit the enter button, your actual input is

20\n

and not just 20. A call to nextInt will now consume the 20 and leave the newline symbol \n in the internal input buffer of System.in. The call to nextLine will now not lead to a new input, since there is still unread input left in System.in. So it will read the \n, leading to an empty input.

So every user input is not only a number, but a full line. As such, it makes much more sense to also use nextLine(), even if reading just an age. The corrected code which works as intended is:

Scanner scanner = new Scanner(System.in);

System.out.println("Enter your age:");
// Now nextLine, not nextInt anymore
int age = Integer.parseInt(scanner.nextLine());
System.out.println("Enter your name:");
String name = scanner.nextLine();

System.out.println("Hello " + name + ", you are " + age + " years old");

The nextXXX methods, such as nextInt can be useful when reading multi-input from a single line. For example when you enter 20 John in a single line.

rough obsidian Nov 22, 2023, 9:59 PM

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This explains the nextInt vs nextLine

#Issue with the Loop