CODEWITHSUNDEEP
java
direwolf96
direwolf96

Reputation: 1

Correction for runtime error while using Scanner in Java

I have an assignment I have to complete for school to make a GPA calculator. It has three classes all together. In the class I am working in, we had to set up multiple methods. I this particular method, the "readGradeCount" method, the user is supposed to enter how many grades they want to calculate for their GPA. The user is supposed to enter a positive number and one that is greater than zero. I have to validate this users input, so I decided to do this by using a loop. When I run the code it gives me the following 

java.util.Scanner[delimiters=\p{javaWhitespace}+][position=0][match valid=false][need input=false][source closed=false][skipped=false][group separator=\,][decimal separator=\.][positive prefix=][negative prefix=\Q-\E][positive suffix=][negative suffix=][NaN string=\Q�\E][infinity string=\Q∞\E]

I have no idea what this means, and I have tried many things to fix it but I can't. This is my code for this method: 

public static int readGradeCount()
{
  int count;
  Scanner keyboard;
  keyboard = new Scanner(System.in);

  do
  {
    System.out.print("How many grades? ");
    while(!keyboard.hasNextInt())
    {
      System.out.print("Bad value- " + keyboard + "\n");
      keyboard.next();
      System.out.print("How many grades? ");

    }
    count = keyboard.nextInt();
  }    
  while (count <= 0);

  return count;
}

The value the user enters is supposed to come after the phrase "Bad value- " but I am getting that line of stuff instead. Any suggestions? Thanks for the help!

Upvotes: 0

Views: 96

Answers (2)

Aimert
Aimert

Reputation: 146

You are printing the scanner and that gives a very weird string. Try to capture the value of the keyboard first and printing that.

while (!keyboard.hasNextInt()) {
    final String value = keyboard.next();
    System.out.print("Bad value- " + value + "\n");
...

Upvotes: 1

Riley Carney
Riley Carney

Reputation: 860

You are printing out the Scanner keyboard in the println statement, try using keyboard.next() in the println and then deleting keyboard.next() in your program.

Upvotes: 2

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