Java: Getting input with Integer.parseInt(sc.nextLine()) and scan.nextInt()

Question

I have a Java program below. At first, I tried to get input n as this

int n = sc.nextInt();

But the output was different than expected. It runs the first iteration without taking the userName. After changing to

int n = Integer.parseInt(sc.nextLine());

It's working fine. What was wrong with "int n = sc.nextInt();"?

public class StringUserNameChecker {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        // Number of usernames you want to enter
        int n = Integer.parseInt(sc.nextLine());
        while (n-- != 0) {
            String userName = sc.nextLine();
            System.out.println("Your username is " + userName);
        }
    }
}

if it's just an integer then `username` becomes nothing because there's nothing left to read (except probably an `\n` or something of that kind) — Shinra tensei, Nov 08 '17 at 09:51

score 0 · Answer 1 · edited May 11 '22 at 18:17

0

When sc.nextLine() is used after sc.nextInt(), it will read a newline character after the integer input.

So, to run your code correctly, you'll have to use sc.nextLine() after sc.nextInt(), like this:

int n = sc.nextInt();
sc.nextLine();
while(n-- > 0) {
    String username = sc.nextLine();
}

edited May 11 '22 at 18:17

Peter Mortensen

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answered Nov 08 '17 at 09:54

Sushant Gaikwad

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1

Correct but badly explained. – AxelH Nov 08 '17 at 10:01

Java: Getting input with Integer.parseInt(sc.nextLine()) and scan.nextInt()

1 Answers1