Sometimes you have to clear the buffer to print the strings by command sc.nextLine();, here is my correct code:- import java.util.Scanner;

public class Solution { public static void main(String[] args) { Scanner sc=new Scanner(System.in); int x=sc.nextInt(); double y = sc.nextDouble(); sc.nextLine(); String s = sc.nextLine();

    System.out.println("String: "+s);
    System.out.println("Double: "+y);
    System.out.println("Int: "+x);

} }

Java8 solution - passes 100% of test cases

import java.util.Scanner;

public class Solution {
    public static void main(String[] args) {
        /* Read input */
        Scanner scan = new Scanner(System.in);
        int i    = scan.nextInt();
        double d = scan.nextDouble();
        scan.nextLine();              // gets rid of the pesky newline
        String s = scan.nextLine();
        scan.close();
        
        /* Print output */
        System.out.println("String: " + s);
        System.out.println("Double: " + d);
        System.out.println("Int: " + i);
    }
}

Invisible "\n" newlines

At the end of each line of the provided input, there is an invisible "\n" which is called a newline character. This basically represents what happens when you press the "Enter" key when typing.

1) The first scan.nextInt() grabs the first "int" from the input. It still leaves the "\n" on that line.

2) The second scan.nextDouble() skips over that "\n" and grabs the first "double" it sees.

3) Now is the tricky part. It turns out scan.nextLine() works differently than scan.nextInt() and scan.nextDouble() in that it doesn't skip over any "\n" newline characters. What it does is it reads and returns all characters until it reaches the "\n" (which it gets rid of), and stops. In the scenario above, since "\n" is the first thing we see, we get an empty String back on our 1st call to scan.nextLine().

4) Our next call to scan.nextLine() will start on line 3 of input and grab the String we want.

Hope this helps.

From my HackerRank Java solutions

I used code below to solved this problem.

import java.util.Scanner;

public class Solution {

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int i = scan.nextInt();
        Double d = scan.nextDouble();
        String s = scan.nextLine();
        s = scan.nextLine();

        System.out.println("String: " + s);
        System.out.println("Double: " + d);
        System.out.println("Int: " + i);
    }
}

You must enter scan.nextLine(); for s. The first for clean buffer or null character from last scanner input operation and this happen because scanner will doing different job for different data type (Numeric to String). The second, scan.nextLine(); ready for capture our String input.

Check me if i wrong. Thanks

Read the note carefully: "If you use the nextLine() method immediately following the nextInt() method, recall that nextInt() reads integer tokens; because of this, the last newline character for that line of integer input is still queued in the input buffer and the next nextLine() will be reading the remainder of the integer line (which is empty)."

It clearly states that After using nextInt() or nextDouble() everytime the cursor should go to next line to do this u should use scannerobj.nextLine()**

*Solution is here *

import java.util.*;

public class Solution {

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


    int i = scan.nextInt();
    double d = scan.nextDouble();
    scan.nextLine();  // jumpt to next line
    String s = scan.nextLine();  // take the whole input and automatically moves the scanner down after returning the current line.
    scan.close();

    // Write your code here.

    System.out.println("String: " + s);
    System.out.println("Double: " + d);
    System.out.println("Int: " + i);
}

}

*Have a nice coding day *