i have a question if u try s = sc.hansNext(); , it ouptputs the 1st word in the next line's string , shouldn't it also read the enter key ( empty output) instead of jumping to the next like nextLine() did in our case.

next() places the cursor in the same line after reading the input.

nextLine() positions the cursor in the next line ater reading output

Thanks for your explaination..BTW what is the keyboard buffer?

how

keyboardbuffer pls explain.... btw great explanation

Definition - What does Keyboard Buffer mean?

A keyboard buffer is a small area in the computer's memory (RAM) that is used to temporarily store the keystrokes from the keyboard before they are processed by the CPU. This is done because there is a delay between the pressing of the key and the sending of the signals, so to avoid timing issues, all the keystrokes are stored in the keyboard buffer until the user presses the "enter" key or a similar command which is very evident in command-line processing or time-sharing systems of generations past. But in today's modern computing environment with fast hardware and more memory, the keyboard buffer is not as obvious. Techopedia explains Keyboard Buffer

The keyboard buffer is used by the operating system to poll key strokes before processing the commands formed by those key presses. This is used to avoid premature processing of invalid commands and to avoid synchronization issues between the user and the computer, since without the buffer, a computer might be expecting a series of key presses from the user that does not come in time. A buffer which stores the typed characters, and essentially the commands, solves this issue of synchronization.

It is also a way to limit the input so that the computer is not flooded with inputs or interrupt requests, especially if a key combination is used for a specific command, such as the ctrl+alt+del command which brings up the task manager. If too many keys are pressed at once, the keyboard buffer returns an error and this is usually heard as a beep generated by the motherboard's built-in speaker. In older machines with slow CPU and RAM, it is possible for the user to type faster than the buffer can store the data, so an error is returned that the keyboard buffer is full. In this case, the user must simply type slower. However, this is no longer a problem in modern computers.

Thank you

What is the reason that it occurs only when we take INT/FLOAT input and then STRING.

For Example,

int i = sc.nextInt();
String s = sc.nextLine();

It will keep key buffer.

But here,

int i = sc.nextInt();
double d = sc.nextDouble();

Here in above case no buffer is created.

thanku sir

thanks buddy

Why does this concept come only while scannig the String and why not while scaning int and double?

could u please eloborate.. I did not quite understand.

Nice explanation. Thanks!

thank you

Then, incase of nextDouble(), it should take "enter key" which has been given after providing int value to nextInt()!!! Why only before nextLine() of String??

thanks

Thank you ! I was confusing with the buffer info.

i didn't get u sir..! please again explain..!!

Thanks it helped me.

can u explain above statement with example

ok but after nextint() why not it effect the double value why it effect only string please explain

Thankyou for a descriptive explanation!! :)

thanks

If I replace String s = scan.nextLine() with String s = scan.next() in the code, then why does it outputs String : welcome. Now why does it kept enter key unread and allow user to input.

Another good explaination & better implementation can be found here:

https://www.geeksforgeeks.org/difference-between-scanner-and-bufferreader-class-in-java/

So I think this challenge was also meant to teach that you will have to get comfortable with looking for helpful solutions to aid in overcoming challenges now and in the furture.

I ended up not even using the Scanner Class altogether but Integer.parseInt() was what was needed regardless.

import java.io.*;

public class Solution {

    public static void main(String[] args) throws IOException{
        BufferedReader scan = new BufferedReader(new InputStreamReader(System.in));
        int i = Integer.parseInt(scan.readLine());
        double d = Double.parseDouble(scan.readLine());
        String s = scan.readLine();

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

So as long as the compiler takes the n as a string.. I guess this implicitly imples that either: 1- The compiler has to be improved (As taking n as a strong sounds illogical & not wise), 2- Java itself has to be modified?

Disclaimer: I'm sorry in advance if these - kind of investigative questions or suggestions - sound naive.. Simply because I'm a beginner, and "can't relate" nor judge if the compiler and Java are doing the right things or not (I guess they're doing the right thing, but I'm not aware of the reason behind it, yet).

My code is same as yours but it is not working for me! It is still throwing me an error.

i have written same but i am getting why will you please help me?

yeah me to code has been nt working

use .nextLine() it will work.

same here

It works to some extent, but not really. It doesn't pass the first test case, but it passes the second.

Why doesn't nextFloat() pass the first test case? Simply because it doesn't generatre the exact expected output. The expected output is (3.1415), and it generates (3.1414999961853027).

Reason? Simply because double is of a bigger size (8 bytes - up to 15 significant decimal digits), while float is smaller (4 bytes - up to 6-7 significant decimal digits).

Reference (I hope this doesn't violate HackerRank policies): http://bit.ly/2DqTF9W

use scan.nextLine before String s=scan.nextLine(); --->scan.nextLine(); --->String s=scan.nextLine(); it will run and u will get final ans

Thank You Gamma2626 for your valuable explanation

So when you call nextDouble() does it continue to search until it finds a double even if there is a new line character?

but then why did scan.nextInt() took 141, it should take \n as its input, since scan.nextDouble took 1.2341 and left \n in buffer.... Please explain??

Thanks Good explanation

but why dont eclipse ask for this....in eclipse my program complied without adding that line???

awesome explanation......thanks

Hey I am using nextLine() after nextInt and then then String input.After that again a nextLine.Then double input.Itsnot working.But the nextLine after each input should clear tge newlines and should work.Then y aint this working?Y are we required to write nextLine after int and double inputs before String input only?

not empty space, but '\n' enter key after int or double

why do you need String s=sc.nextLine(); after reading the String?

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

    int x=sc.nextInt();
    double y =sc.nextDouble();
    String e= sc.nextLine();

    String s=sc.nextLine();





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

}

can you please explain, why should we use String s =sc.nextLin();

Thank you arivero5

    Scanner scan = new Scanner(System.in);

    int i = scan.nextInt();

    double d=scan.nextDouble();

// The next two line will solve the problem to print stringline

    scan.nextLine();
    String s=scan.nextLine();

    // Write your code here.

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

yep you are correct but following code run in the command prompt properly though I didn't use another 'String e' to the programm. can you explain why.

import java.util.Scanner;

public class Solution {

public static void main(String[] args) {
    Scanner scan = new Scanner(System.in);
    Scanner scan2 = new Scanner(System.in);

    int i = scan.nextInt(); 
    double d = scan.nextDouble();
    String s=scan2.nextLine();

    // Write your code here.

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

}

Yes it is a class

It is not mandatory, nevertheless, it is a good practice when programming that became a convention for programmers. It makes your code more clear, elegant, and readable.

To the extent that in some IDEs, or maybe platforms, if you start your class name with capital letter it'll automatically be colored in a special color; and thus visually distinguished from the rest of your code. Example: Check the first answer of this question - http://bit.ly/2C0v9hQ

This is a bit late but if you are still having questions regarding this then I will give you an answer. Using only "scan.nextInt()" causes the program to search for an "int". If it finds it, the code will still remain on the same line while executing the next command which is "scan.nextDouble". mainrajeev's code is a design specifically for this requirement. Since you can predict what will appear on every line, it is better for the Scanner to retrieve each line and parse the required integer, double, and string.

in this code first two cases are executing properly but not others

Who said you can't? Both codes worked properly, & passed the test cases without errors. So recheck your syntax and have a look at your code again to verify that you've written it the same way as above.

good

scan.nextDouble() skips over all characters (including whitespaces characters such as \n) until it finds a double, so that first extra \n is discarded.

HackerRank solutions.

thanku

i did not create it. it's the second test case when you submit the code.

Not sure. I recommend always using Java 8.