Scanner class always introduces the iteration 0 as an empty string

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I think my code works, but it doesn't let the user to introduce the word 0, automatically fills it with an empty string "". I have realised that if I removed the scan class from reading n and m and hardcode the values n=6, m=2. Then the code works as expected. Anyvbody knows why?

package com.codegym.task.task07.task0720;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Iterator;
import java.util.Scanner;

/*
Shuffled just in time

*/

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

        //write your code here

        int n = sc.nextInt();
        int m = sc.nextInt();

        ArrayList<String> sensei = new ArrayList<String>();

        System.out.println("Write vector:");
        for (int i = 0; i < n; i++) {
            System.out.println("Write word: " + i);
            sensei.add(sc.nextLine());
        }

        for (int i = 0; i < m; i++) {
            sensei.add(sensei.get(i));
        }

        for (int i = 0; i < m; i++) {
            sensei.remove(0);
        }

        for (int i = 0; i < sensei.size(); i++) {
            System.out.println(sensei.get(i));
        }

    }
}

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Ricardo

Nivel 10

8 agosto 2021, 09:38

I must say I did solve this problem by adding in line 23:

sc.nextLine();

However, I don't know why I need to do that and would like if someone can explain it to me. My guess is that for any unknown reason, scan doesn't like jumping from reading integers to read Strings.

Lisa

Nivel 41

8 agosto 2021, 12:08

yuppidu... dumdidum... that's easy... 🧐 the Scanner class breaks the input into tokens using a delimiter and that's whitespace by default. The linebreak is such a whitespace. Look at the below code. next() or nextInt() first remove whitespace till another character is found. Then they grab everyting till the next whitespace (here a linebreak) and that next whitespace is left in the pipe. Means the first nextInt reads till the \n linebreak (whitspace) leaving the linebreak inside the buffer. The next nextInt skips leading whitspace and reads the 2, the third the 3, again leaving the \n in the pipe. If you now call a nextLine(), it'll see the linebreak \n and reads of course everything up to that linebreak and that's an empty line and not 'yeah yeah' as you might expect. Another readLine then reads that entire line up to and including the linebreak removing it from the buffer. Now the yeah yeah... that's the entire miracle 🧙‍♂️

Scanner sc = new Scanner("  1\n2 3\nyeah yeah\nnot read content");
System.out.println(sc.nextInt());
System.out.println(sc.nextInt());
System.out.println(sc.nextInt());
System.out.println(sc.nextLine()); // emtpy line, just the \n
System.out.println(sc.nextLine()); // reading yeah yeah

If you understood this, you are also crazy, yeah yeah 😜🤪 Doesn't matter at all, I'm simply a genius...🧠 but a lovable one 😍, just admit it...

Ricardo

Nivel 10

11 agosto 2021, 21:19

Thank you lovable genius. So every time the sc class ask for a new Int, it records the Int and a linebreak \n. If I understand you properly, a linebreak \n it is a combination of a whitespace and a line break (ASCII 32 + ASCII 10) in this order (Now this is the part I don't fully understand). So the Next Line reads the second part of the linebreak (ASCII 10) and calls it a Line, thus my empty line.

Lisa

Nivel 41

12 agosto 2021, 12:50

not really... unfortunately. I'd love I to bring you better news 😥 Let's try again. nextint reads from the input till it finds some int chars and stops when it finds the first whitespace (including linebreaks). It doesn't add that whitespace to the int. It is still in the pipe Example

Scanner sc = new Scanner("1 2\n");
System.out.println(sc.nextInt());
System.out.println(sc.nextLine());

nextInt reads the 1 and stops right before the space. The following nextLine reads the space, the 2 and the linebreak. If you had as input just "1\n" then the same happens. nextInt reads the 1, finds a whitespace (the \n, linebreak) and stops leaving the linebreak in the buffer. The next readLine just jas a linebreak to read. Poor little thing. That's what you experienced. You had to clear the input buffer from the linebreak. Yeah, yeah 😜🤪, I bet you're the Scanner champ now 🥳🥂

Ricardo

Nivel 10

14 agosto 2021, 09:41

Great, that is very useful. I know understand how it works. The last question I have. Is there any specific command to clean the pipeline? I know I can do sc.NextLine(), but that is rather cryptic and I will have to add a comment saying, this is for cleaning the pipeline. I would rather user a specific command for cleaning it if such a command exist. You are amazing Lisa, thank you so much!!

Lisa

Nivel 41

14 agosto 2021, 19:33

Not that I know of. And that's just how the scanner class works. The function of the nextLine methods is to read everything up to and including the next line break. So the purpose is to clear the buffer. Maybe you could abuse the skip method, like this

sc.skip(".*\n");

However I'm nearly sure nextLine works more reliable 😏 Oh... and thanks for your kind words 😊 blush, blush... yeah, yeah 😜🤪