Input mismatch error

Question

So, I have this little program here and I've been getting the input mismatch error. I'll post the error message below the code. To be honest, I have 0 clue why the error appears. I've checked that all the required variables have correct type assigned. If I understand correctly, the said error happens when you try to input something that is not expected, right? ( double instead of int or whatever). I probably did some stupid mistake somewhere that I cant see for some reason, so I'd appreciate your help.

    Scanner sc = new Scanner(System.in);

    int enEur = 0;
    int dvaEur = 0;
    boolean bankrot = false;
    int placilo;


    while (sc.hasNextInt() || bankrot == false){
        placilo = sc.nextInt();
         if(placilo == 1){
             enEur += placilo;
         }
         else{
             enEur --;
             bankrot = test(enEur);
             dvaEur ++;
         }
     }

So, this is the content of my main() method. The error appears in the line: placilo = sc.nextInt(). Here's the copy of the said error:

    Exception in thread "main" java.util.InputMismatchException
    at java.util.Scanner.throwFor(Scanner.java:864)
    at java.util.Scanner.next(Scanner.java:1485)
    at java.util.Scanner.nextInt(Scanner.java:2117)
    at java.util.Scanner.nextInt(Scanner.java:2076)
    at evroSop.main(evroSop.java:16)

The input that I put in is basically a line of separate integers like: 1 1 1 1 2 1... and so on. I've got different examples, all of which are lines of integers of different lengths.

Cherio, George

you need to get the carriage return by calling `nextLine` after each call of `nextInt`. Otherwise you will try to input `\n` into an int, which will cause you the error — SomeJavaGuy, Oct 13 '15 at 13:55
@KevinEsche Hmm.. interesting, I need to make a note of that. However, I've put the `sc.nextLine()` below the `sc.nextInt()`, but nothing changes :'( — George Waat, Oct 13 '15 at 14:02

score 1 · Accepted Answer · answered Oct 13 '15 at 14:12

1

The condition while (sc.hasNextInt() || bankrot == false) means that you'll continue trying to read from the scanner as long as there is more input or bankrot is still false. So, if you run out of inputs and still aren't bankrupt, you'll get an exception because you'll try reading next input even though there is no more of it.

The condition should be and, not or, probably.

answered Oct 13 '15 at 14:12

Jiri Tousek

12,211
5
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You sir, are my savior. Dunno why I never thought of that, I think I need to clear some info about what the and and or really mean and how it all works. This solved my problem completely, thank you. – George Waat Oct 13 '15 at 14:20

Ravindra babu · Answer 2 · 2015-10-13T14:13:40.207

0

Read the line as String and split it into int[]

after hasNext() call,

String str = sc.nextLine();

and split this String into int[]

Example code:

        Scanner sc = new Scanner(System.in);
        while(sc.hasNext()){
                String str = sc.nextLine();
                String strArray[] = str.split(" ");
                int intArray[] = new int[strArray.length];

               for (int i = 0; i < intArray.length ; i++) {
                   intArray[i] = Integer.parseInt(strArray[i]);
               }

               for (int j : intArray) {
                    System.out.println(j);
               }
        }

Input:

1 123 456

Output:

1

123

456

edited Oct 13 '15 at 14:13

answered Oct 13 '15 at 14:05

Ravindra babu

37,698
11
250
211

While this is reasonable advice, my professor will strangle me if I get anywhere close to transforming strings to ints and vice versa. For some reason we are not allowed to do that. But thanks for the advice non the less. – George Waat Oct 13 '15 at 14:19

Input mismatch error

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