Home/ Questions/Error: exception in thread "main" java.util.inputmismatchexception - Quick tips
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Inès Renault
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Asked: 2022-05-11T18:37:13+00:00 In:

Error: exception in thread “main” java.util.inputmismatchexception – Quick tips

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This is the program I run:

import java.util.Scanner;
 public class Sal {
 
  public static void main (String args []){ 
  Scanner s=new Scanner(System.in);
  System.out.println("Enter 10 areas and their pincodes");
  String area[]=new String [10];
  int pincode[]=new int [10];
  String search;
  int chk=0;
  int p=0;
 
  for (int i=0;i<=9;i++){
  area[i]=s.nextLine();
  pincode[i]=s.nextInt();
  }
 
  System.out.println("Enter Search"); 
  search=s.nextLine();
 
  for (int j=0;j<=9;j++){
  if(search==area[j]){
  chk=1;
  j=p;
  break;
  }
  }
 
  if(chk==1){
  System.out.println("Search Found "+"Pincode : "+pincode[p] );
  } else {
  System.out.println("Search not Found");
  }
  }
 }

After I run, it returns an error

Exception in thread "main" java.util.InputMismatchException
 at java.util.Scanner.throwFor(Unknown Source)
 at java.util.Scanner.next(Unknown Source)
 at java.util.Scanner.nextInt(Unknown Source)
 at java.util.Scanner.nextInt(Unknown Source)
 at Sal.main(Sal.java:14)

Does anyone have any suggestions for my problem: exception in thread “main” java.util.inputmismatchexception in the programs – How to correct it?

exception in thread "main"
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2 Answers

  1. Best Answer
    Expert

    The cause: The input does not match the Integer regular expression, or is out of range. I guess that at some point, your call to .nextLine() swallows up an entire line, and the next line starts with an “area”.

    Solution: 

    for (int i=0;i<=9;i++){

area[i]=s.next();// Use this for String Input

pincode[i]=s.nextInt();

s.nextLine();//Use this for going to next line of input

}

    Compare Strings with the compareTo method (which returns zero if they are equal)

    if(search.compareTo(area[j])==0){
    • 21

  2. The Scanner#nextInt() docs:

    InputMismatchException – if the next token does not match the Integer regular expression, or is out of range

    It sounds like your Scanner has tried to read in a int, but it is getting something it cannot turn into a int (either the number it read isn’t readable or it is too large).

    This is the function you call: 

    for (int i=0;i<=9;i++){
 area[i]=s.nextLine();
 pincode[i]=s.nextInt(); // <-- the culprit
}

    I suspect that .nextLine()‘s call takes up a whole line. The next line begins with an “area”. Without knowing what format you want the input to look like, I can’t do much more.

    • 15

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