Scanner is skipping nextLine() after using next() or nextFoo()?

Question

I am using the Scanner methods nextInt() and nextLine() for reading input.

It looks like this:

System.out.println("Enter numerical value");    
int option;
option = input.nextInt(); // Read numerical value from input
System.out.println("Enter 1st string"); 
String string1 = input.nextLine(); // Read 1st string (this is skipped)
System.out.println("Enter 2nd string");
String string2 = input.nextLine(); // Read 2nd string (this appears right after reading numerical value)

The problem is that after entering the numerical value, the first input.nextLine() is skipped and the second input.nextLine() is executed, so that my output looks like this:

Enter numerical value
3   // This is my input
Enter 1st string    // The program is supposed to stop here and wait for my input, but is skipped
Enter 2nd string    // ...and this line is executed and waits for my input

I tested my application and it looks like the problem lies in using input.nextInt(). If I delete it, then both string1 = input.nextLine() and string2 = input.nextLine() are executed as I want them to be.

Fibrovascular

Or you could be like me and use BufferedReader :) I don't care if it's old school, it has always worked and always will work for me. Also, knowledge of BufferedReader has application elsewhere. I simply don't like Scanner. – Doughnut

Answer 1

That's because the Scanner.nextInt method does not read the newline character in your input created by hitting "Enter," and so the call to Scanner.nextLine returns after reading that newline.

You will encounter the similar behaviour when you use Scanner.nextLine after Scanner.next() or any Scanner.nextFoo method (except nextLine itself).

Workaround:

• Either put a Scanner.nextLine call after each Scanner.nextInt or Scanner.nextFoo to consume rest of that line including newline

  int option = input.nextInt();
  input.nextLine();  // Consume newline left-over
  String str1 = input.nextLine();
  

• Or, even better, read the input through Scanner.nextLine and convert your input to the proper format you need. For example, you may convert to an integer using Integer.parseInt(String) method.

  int option = 0;
  try {
      option = Integer.parseInt(input.nextLine());
  } catch (NumberFormatException e) {
      e.printStackTrace();
  }
  String str1 = input.nextLine();
  

Milewski

Couldn't I use only option = Integer.parseInt(input.nextLine());} instead of full try{...}? Is there any weakness of using only above? – Fibrovascular

@blekione. You have to use try-catch, because Integer.parseInt throws NumberFormatException when an invalid argument is passed to it. You will learn about exception later on. For E.G: - Integer.parseInt("abc"). You don't want "abc" to get converted to int right? – Milewski

@blekione. So, in the above case, your code will halt at that point, and you won't be able to continue the execution. With Exception Handling, you can handle such kind of conditions. – Milewski

To which extent is the latter better? AFAIK Scanner#nextInt() is way more lenient in finding correct ints, by allowing group commas and locale prefixes and suffixes. Integer#parseInt() allows digits and decimal point only plus an optional sign. – Mi

I personally prefer a Scanner#hasNextFoo check beforehand instead of a try-catch, but that works too. – Priedieu

Scanner scan = new Scanner(System.in); int i = scan.nextInt(); Double d = scan.nextDouble(); int option = 0; try { option = Integer.parseInt(scan.nextLine()); } catch (NumberFormatException e) { //e.printStackTrace(); } String str = scan.nextLine(); System.out.println("String: " + str); System.out.println("Double: " + d); System.out.println("Int: " + i); – Vicinage

Use ParseDouble has a new problem, that nextDouble use the regional configuration of decimal (. or ,) but Parsedouble always receive US decimal ( . ). – Pilkington

Another solution could be to use a delimiter that will also ignore newlines besides just spaces. e.g. Scanner input = new Scanner(System.in).useDelimiter("\\s+"); – Farcy

input.nextLine() not working for me. I am surprised because I have used it before. String s=input().nextLine() works. I don't know why it is demanding a String object. – Liva

A very important issue for new Java programmers and Excellent presentation about the issue. – Jehad

If that is the case, Why don't we encounter the same problem if we use Scanner.next() call after Scanner.nextInt(). Why don't I get a blank string in Scanner.next() ? – Trinatrinal

thankyou so much for the answer! I am working around my program since an hour but can't figure out the bug. Your first point conveys better than the second according to me @RohitJain – Fabulous

Answer 2

The problem is with the input.nextInt() method; it only reads the int value. So when you continue reading with input.nextLine() you receive the "\n" Enter key. So to skip this you have to add the input.nextLine().

Try it like this, instead:

System.out.print("Insert a number: ");
int number = input.nextInt();
input.nextLine(); // This line you have to add (It consumes the \n character)
System.out.print("Text1: ");
String text1 = input.nextLine();
System.out.print("Text2: ");
String text2 = input.nextLine();

Horntail

FYI: merged from #7057249 – Hemo

If that is the case, Why don't we encounter the same problem if we use Scanner.next() call after Scanner.nextInt(). Why don't I get a blank string in Scanner.next() ? – Trinatrinal

Answer 3

It's because when you enter a number then press Enter, input.nextInt() consumes only the number, not the "end of line". When input.nextLine() executes, it consumes the "end of line" still in the buffer from the first input.

Instead, use input.nextLine() immediately after input.nextInt()

Mistake

@Victor it's not bug. it's working as specified. you could argue though that there should be an easy way to tell the api to also consume any whitespace following the target input. – Mistake

I see. thanks @Bohemian, that exactly what i am arguing. I think that this should be changed, perhaps you can point me where to suggest this "issue" in the next JCP. – Bretbretagne

@victor You can request (and find out how to contribute code to) a feature via the Report a Bug or Request a Feature page. – Mistake

FYI: merged from #7057249 – Hemo

If that is the case, Why don't we encounter the same problem if we use Scanner.next() call after Scanner.nextInt(). Why don't I get a blank string in Scanner.next() ? – Trinatrinal

Answer 4

There seem to be many questions about this issue with java.util.Scanner. I think a more readable/idiomatic solution would be to call scanner.skip("[\r\n]+") to drop any newline characters after calling nextInt().

EDIT: as @PatrickParker noted below, this will cause an infinite loop if user inputs any whitespace after the number. See their answer for a better pattern to use with skip: https://mcmap.net/q/36338/-string-formatting-in-java-scanner-class

Nonfulfillment

FYI: merged from #7057249 – Hemo

I know what we do to remove the data in buffer, but this case, please help me with this: https://mcmap.net/q/36184/-scanner-is-skipping-nextline-after-using-next-or-nextfoo – Energetics

FYI, this will cause an infinite loop is the user types any tabs or spaces after the numeric input. See my answer here for a better skip: https://mcmap.net/q/36338/-string-formatting-in-java-scanner-class – Planimetry

@PatrickParker: it does indeed cause an infinite loop! Thanks, edited the answer to link to yours. – Nonfulfillment

Answer 5

TL;DR
nextLine() is safe to call when (a) it is first reading instruction, (b) previous reading instruction was also nextLine().

If you are not sure that either of above is true you can use scanner.skip("\\R?") before calling scanner.nextLine() since calls like next() nextInt() will leave potential line separator - created by return key which will affect result of nextLine(). The .skip("\\R?") will let us consume this unnecessary line separator.

skip uses regex where

• \R represents line separators
• ? will make \R optional - which will prevent skip method from:
  • waiting for matching sequence
    • in case of reaching end of still opened source of data like System.in, input stream from socket, etc.
  • throwing java.util.NoSuchElementException in case of
    • terminated/closed source of data,
    • or when existing data doesn't match what we want to skip

Things you need to know:

• text which represents few lines also contains non-printable characters between lines (we call them line separators) like

• carriage return (CR - in String literals represented as "\r")

• line feed (LF - in String literals represented as "\n")

• when you are reading data from the console, it allows the user to type his response and when he is done he needs to somehow confirm that fact. To do so, the user is required to press "enter"/"return" key on the keyboard.

What is important is that this key beside ensuring placing user data to standard input (represented by System.in which is read by Scanner) also sends OS dependant line separators (like for Windows \r\n) after it.

So when you are asking the user for value like age, and user types 42 and presses enter, standard input will contain "42\r\n".

Problem

Scanner#nextInt (and other Scanner#nextType methods) doesn't allow Scanner to consume these line separators. It will read them from System.in (how else Scanner would know that there are no more digits from the user which represent age value than facing whitespace?) which will remove them from standard input, but it will also cache those line separators internally. What we need to remember, is that all of the Scanner methods are always scanning starting from the cached text.

Now Scanner#nextLine() simply collects and returns all characters until it finds line separators (or end of stream). But since line separators after reading the number from the console are found immediately in Scanner's cache, it returns empty String, meaning that Scanner was not able to find any character before those line separators (or end of stream).
BTW nextLine also consumes those line separators.

Solution

So when you want to ask for number and then for entire line while avoiding that empty string as result of nextLine, either

• consume line separator left by nextInt from Scanners cache by
• calling nextLine,
• or IMO more readable way would be by calling skip("\\R") or skip("\r\n|\r|\n") to let Scanner skip part matched by line separator (more info about \R: https://stackoverflow.com/a/31060125)
• don't use nextInt (nor next, or any nextTYPE methods) at all. Instead read entire data line-by-line using nextLine and parse numbers from each line (assuming one line contains only one number) to proper type like int via Integer.parseInt.

---

BTW: Scanner#nextType methods can skip delimiters (by default all whitespaces like tabs, line separators) including those cached by scanner, until they will find next non-delimiter value (token). Thanks to that for input like "42\r\n\r\n321\r\n\r\n\r\nfoobar" code

int num1 = sc.nextInt();
int num2 = sc.nextInt();
String name = sc.next();

will be able to properly assign num1=42 num2=321 name=foobar.

Desantis

Answer 6

It does that because input.nextInt(); doesn't capture the newline. you could do like the others proposed by adding an input.nextLine(); underneath.
Alternatively you can do it C# style and parse a nextLine to an integer like so:

int number = Integer.parseInt(input.nextLine()); 

Doing this works just as well, and it saves you a line of code.

Lipcombe

FYI: merged from #7057249 – Hemo

you have to use try catch here. What will happen if the input is not a number. NumberFormatException need to be handled here. – Tonicity

This assumes that there is only one int token on each line you try this for. – Fruma

If that is the case, Why don't we encounter the same problem if we use Scanner.next() call after Scanner.nextInt(). Why don't I get a blank string in Scanner.next() ? – Trinatrinal

@Trinatrinal I wish I remembered, I haven't touched Java since I wrote that answer in '13 – Lipcombe

Answer 7

Instead of input.nextLine() use input.next(), that should solve the problem.

Modified code:

public static Scanner input = new Scanner(System.in);

public static void main(String[] args)
{
    System.out.print("Insert a number: ");
    int number = input.nextInt();
    System.out.print("Text1: ");
    String text1 = input.next();
    System.out.print("Text2: ");
    String text2 = input.next();
}

Overcurious

FYI: merged from #7057249 – Hemo

Answer 8

If you want to read both strings and ints, a solution is to use two Scanners:

Scanner stringScanner = new Scanner(System.in);
Scanner intScanner = new Scanner(System.in);

intScanner.nextInt();
String s = stringScanner.nextLine(); // unaffected by previous nextInt()
System.out.println(s);

intScanner.close();
stringScanner.close();

Gissing

please do not do this. Just one scanner per inputstream. Also, when you do intScanner.close(); you close stringScanner + the whole System.in for the rest of your execution. – Seattle

Answer 9

In order to avoid the issue, use nextLine(); immediately after nextInt(); as it helps in clearing out the buffer. When you press ENTER the nextInt(); does not capture the new line and hence, skips the Scanner code later.

Scanner scanner =  new Scanner(System.in);
int option = scanner.nextInt();
scanner.nextLine(); //clearing the buffer

Mussolini

Answer 10

If you want to scan input fast without getting confused into Scanner class nextLine() method , Use Custom Input Scanner for it .

Code :

class ScanReader {
/**
* @author Nikunj Khokhar
*/
    private byte[] buf = new byte[4 * 1024];
    private int index;
    private BufferedInputStream in;
    private int total;

    public ScanReader(InputStream inputStream) {
        in = new BufferedInputStream(inputStream);
    }

    private int scan() throws IOException {
        if (index >= total) {
            index = 0;
            total = in.read(buf);
            if (total <= 0) return -1;
        }
        return buf[index++];
    }
    public char scanChar(){
        int c=scan();
        while (isWhiteSpace(c))c=scan();
        return (char)c;
    }


    public int scanInt() throws IOException {
        int integer = 0;
        int n = scan();
        while (isWhiteSpace(n)) n = scan();
        int neg = 1;
        if (n == '-') {
            neg = -1;
            n = scan();
        }
        while (!isWhiteSpace(n)) {
            if (n >= '0' && n <= '9') {
                integer *= 10;
                integer += n - '0';
                n = scan();
            }
        }
        return neg * integer;
    }

    public String scanString() throws IOException {
        int c = scan();
        while (isWhiteSpace(c)) c = scan();
        StringBuilder res = new StringBuilder();
        do {
            res.appendCodePoint(c);
            c = scan();
        } while (!isWhiteSpace(c));
        return res.toString();
    }

    private boolean isWhiteSpace(int n) {
        if (n == ' ' || n == '\n' || n == '\r' || n == '\t' || n == -1) return true;
        else return false;
    }

    public long scanLong() throws IOException {
        long integer = 0;
        int n = scan();
        while (isWhiteSpace(n)) n = scan();
        int neg = 1;
        if (n == '-') {
            neg = -1;
            n = scan();
        }
        while (!isWhiteSpace(n)) {
            if (n >= '0' && n <= '9') {
                integer *= 10;
                integer += n - '0';
                n = scan();
            }
        }
        return neg * integer;
    }

    public void scanLong(long[] A) throws IOException {
        for (int i = 0; i < A.length; i++) A[i] = scanLong();
    }

    public void scanInt(int[] A) throws IOException {
        for (int i = 0; i < A.length; i++) A[i] = scanInt();
    }

    public double scanDouble() throws IOException {
        int c = scan();
        while (isWhiteSpace(c)) c = scan();
        int sgn = 1;
        if (c == '-') {
            sgn = -1;
            c = scan();
        }
        double res = 0;
        while (!isWhiteSpace(c) && c != '.') {
            if (c == 'e' || c == 'E') {
                return res * Math.pow(10, scanInt());
            }
            res *= 10;
            res += c - '0';
            c = scan();
        }
        if (c == '.') {
            c = scan();
            double m = 1;
            while (!isWhiteSpace(c)) {
                if (c == 'e' || c == 'E') {
                    return res * Math.pow(10, scanInt());
                }
                m /= 10;
                res += (c - '0') * m;
                c = scan();
            }
        }
        return res * sgn;
    }

}

Advantages :

• Scans Input faster than BufferReader
• Reduces Time Complexity
• Flushes Buffer for every next input

Methods :

• scanChar() - scan single character
• scanInt() - scan Integer value
• scanLong() - scan Long value
• scanString() - scan String value
• scanDouble() - scan Double value
• scanInt(int[] array) - scans complete Array(Integer)
• scanLong(long[] array) - scans complete Array(Long)

Usage :

1. Copy the Given Code below your java code.
2. Initialise Object for Given Class

ScanReader sc = new ScanReader(System.in); 3. Import necessary Classes :

import java.io.BufferedInputStream; import java.io.IOException; import java.io.InputStream; 4. Throw IOException from your main method to handle Exception 5. Use Provided Methods. 6. Enjoy

Example :

import java.io.BufferedInputStream;
import java.io.IOException;
import java.io.InputStream;
class Main{
    public static void main(String... as) throws IOException{
        ScanReader sc = new ScanReader(System.in);
        int a=sc.scanInt();
        System.out.println(a);
    }
}
class ScanReader....

Entremets

Answer 11

sc.nextLine() is better as compared to parsing the input. Because performance wise it will be good.

Encephalogram

Answer 12

I guess I'm pretty late to the party..

As previously stated, calling input.nextLine() after getting your int value will solve your problem. The reason why your code didn't work was because there was nothing else to store from your input (where you inputted the int) into string1. I'll just shed a little more light to the entire topic.

Consider nextLine() as the odd one out among the nextFoo() methods in the Scanner class. Let's take a quick example.. Let's say we have two lines of code like the ones below:

int firstNumber = input.nextInt();
int secondNumber = input.nextInt();

If we input the value below (as a single line of input)

54 234

The value of our firstNumber and secondNumber variable become 54 and 234 respectively. The reason why this works this way is because a new line feed (i.e \n) IS NOT automatically generated when the nextInt() method takes in the values. It simply takes the "next int" and moves on. This is the same for the rest of the nextFoo() methods except nextLine().

nextLine() generates a new line feed immediately after taking a value; this is what @RohitJain means by saying the new line feed is "consumed".

Lastly, the next() method simply takes the nearest String without generating a new line; this makes this the preferential method for taking separate Strings within the same single line.

I hope this helps.. Merry coding!

Myocardium

what is a "new line feed"? – Aceto

@Abcd A new line feed basically means 'starting from a new line'. – Myocardium

That "54 234" example really clarifies things. – Chungchungking

Answer 13

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

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

Kingsley

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). – Kingsley

Answer 14

if I expect a non-empty input

avoids:
–  loss of data if the following input is eaten by an unchecked scan.nextLine() as workaround
–  loss of data due to only partially read lines because scan.nextLine() was replaced by scan.next() (enter: "yippie ya yeah")
–  Exceptions that are thrown when parsing input with Scanner methods (read first, parse afterwards)

public static Function<Scanner,String> scanLine = (scan -> {
    String s = scan.nextLine();
    return( s.length() == 0 ? scan.nextLine() : s );
  });

used in above example:

System.out.println("Enter numerical value");    
int option = input.nextInt(); // read numerical value from input

System.out.println("Enter 1st string"); 
String string1 = scanLine.apply( input ); // read 1st string
System.out.println("Enter 2nd string");
String string2 = scanLine.apply( input ); // read 2nd string

Arnettaarnette

never seen this before. How would this behave in (beware spam) the close scenario? https://mcmap.net/q/36339/-should-a-scanner-only-be-instantiated-only-once-if-that-39-s-the-case-why-so – Seattle

@Seattle While here the problem is, to prevent an empty input (working in any scenario), is the other problem mentioned above, that the resources of an active Scanner are being released through of a wrong placed call of close. The use of autoclosing is the correct way to close the Scanner or any other AutoCloseable. – Arnettaarnette

Answer 15

In one of my usecase, I had the scenario of reading a string value preceded by a couple of integer values. I had to use a "for / while loop" to read the values. And none of the above suggestions worked in this case.

Using input.next() instead of input.nextLine() fixed the issue. Hope this might be helpful for those dealing with similar scenario.

Maier

Short answer is that whitespace matters. If you have multiple integers on the same line, you can still use nextLine and split them, and you still need a nextLine after the last next call on the line of input – Seringa

Answer 16

Use 2 scanner objects instead of one

Scanner input = new Scanner(System.in);
System.out.println("Enter numerical value");    
int option;
Scanner input2 = new Scanner(System.in);
option = input2.nextInt();

Bissau

Answer 17

As nextXXX() methods don't read newline, except nextLine(). We can skip the newline after reading any non-string value (int in this case) by using scanner.skip() as below:

Scanner sc = new Scanner(System.in);
int x = sc.nextInt();
sc.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
System.out.println(x);
double y = sc.nextDouble();
sc.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
System.out.println(y);
char z = sc.next().charAt(0);
sc.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
System.out.println(z);
String hello = sc.nextLine();
System.out.println(hello);
float tt = sc.nextFloat();
sc.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
System.out.println(tt);

Sordid

Answer 18

Use this code it will fix your problem.

System.out.println("Enter numerical value");    
int option;
option = input.nextInt(); // Read numerical value from input
input.nextLine();
System.out.println("Enter 1st string"); 
String string1 = input.nextLine(); // Read 1st string (this is skipped)
System.out.println("Enter 2nd string");
String string2 = input.nextLine(); // Read 2nd string (this appears right after reading numerical value)

Alver

While this solves the problem, it doesn't add any more detail than the original answers or explain why it works – Seringa

Answer 19

To resolve this problem just make a scan.nextLine(), where scan is an instance of the Scanner object. For example, I am using a simple HackerRank Problem for the explanation.

package com.company;
import java.util.Scanner;

public class hackerrank {
public static void main(String[] args) {
    Scanner scan = new Scanner(System.in);
    int i = scan.nextInt();
    double d = scan.nextDouble();
    scan.nextLine(); // This line shall stop the skipping the nextLine() 
    String s = scan.nextLine();
    scan.close();



    // Write your code here.

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

}

Officious

Answer 20

The nextLine() will read enter directly as an empty line without waiting for the text.

Simple solution by adding an extra scanner to consume the empty line:

System.out.println("Enter numerical value");    
int option;
option = input.nextInt(); // Read numerical value from input
input.nextLine();
System.out.println("Enter 1st string"); 
String string1 = input.nextLine(); // Read 1st string (this is skipped)
System.out.println("Enter 2nd string");
String string2 = input.nextLine(); // Read 2nd string (this appears right after reading numerical value)

Hurley

Answer 21

This is a very basic problem for beginner coders in java. The same problem I also have faced when I started java (Self Taught). Actually, when we take an input of integer dataType, it reads only integer value and leaves the newLine(\n) character and this line(i.e. leaved new line by integer dynamic input )creates the problem when we try to take a new input. eg. Like if we take the integer input and then after try to take an String input.

value1=sc.nextInt();
value2=sc.nextLine();

the value2 will auto read the newLine character and will not take the user input.

Solution: just we need to add one line of code before taking the next user input i.e.

sc.nextLine();

or

value1=sc.nextInt();
sc.nextLine();
value2=sc.nextLine();

Note: don't forget to close the Scanner to prevent memory leak;

Mccormac

Answer 22

The problem is with the input.nextInt() method - it only reads the int value. So when you continue reading with input.nextLine() you receive the "\n" Enter key. So to skip this you have to add the input.nextLine(). Hope this should be clear now.

Try it like that:

System.out.print("Insert a number: ");
int number = input.nextInt();
input.nextLine(); // This line you have to add (It consumes the \n character)
System.out.print("Text1: ");
String text1 = input.nextLine();
System.out.print("Text2: ");
String text2 = input.nextLine();

Nadabus

Answer 23

 Scanner scan = new Scanner(System.in);
    int i = scan.nextInt();
    scan.nextLine();//to Ignore the rest of the line after  (integer input)nextInt()
    double d=scan.nextDouble();
    scan.nextLine();
    String s=scan.nextLine();
    scan.close();
    System.out.println("String: " + s);
    System.out.println("Double: " + d);
    System.out.println("Int: " + i);

Heterography

Answer 24

Use BufferedReader class to input string, this will not create problems

Jaffe

Answer 25

Why not use a new Scanner for every reading? Like below. With this approach you will not confront your problem.

int i = new Scanner(System.in).nextInt();

Spieler

But then you have to close the Scanner to prevent memory leak. Wasting time? – Ceremonious

Doesn't the GC take care of that if the Scanner is wrapped in a method? Like: String getInput() {return new Scanner(System.in).nextLine()}; – Spieler

This is absolutely incorrect. nextInt() won't consume the newline, regardless of whether it's in a "new" Scanner or an already used one. – Burglarious