CODEWITHSUNDEEP
java
Bob G.
Bob G.

Reputation: 143

Java-Try and Catch Statement

I am just starting to write a blackjack game with java. I am trying to get the program to ask the user to enter again if the cash they typed in to start with isn't a valid integer. I see many examples of the try statement with catch, but none of them is working. The program gives the error InputMismatchException cannot be resolved to a type. A thread I have followed is this one, and I have the exact same code, just different variable name. Here it is. Java InputMismatchException

Here is my code:

        Scanner input_var=new Scanner(System.in);
        System.out.println("Welcome to BlackJack!");
        System.out.println("Enter how much money you will start with");
        System.out.println("Starting cash must be whole number");
        int money=0;
        do{

        try{
            System.out.println("Enter how much money you will start with: ");
            money=input_var.nextInt();
            }

        catch (InputMismatchException e){
            System.out.println("Sry, not a valid integer");
            input_var.nextInt();
        }
        input_var.nextLine();
        }while (money<=0);

Any help with why my almost exact code isn't working would be greatly appreciated. Thanks for your time and effort.

Upvotes: 0

Views: 277

Answers (4)

dumbPotato21
dumbPotato21

Reputation: 5695

Consume using input.next(), as input.nextInt() doesn't exist. That's the whole point of the Exception.

do{
    try{
        System.out.println("Enter how much money you will start with: ");
        money=input_var.nextInt();
    }catch (InputMismatchException e){
        System.out.println("Sry, not a valid integer");
        input_var.next();
    }
}while (money<=0);

Upvotes: 2

nixin72
nixin72

Reputation: 342

Maybe you're looking for NumberFormatExcpetion? It's what gets thrown when converting strings to numbers. Try doing this instead: 

    Scanner input_var=new Scanner(System.in);
    System.out.println("Welcome to BlackJack!");
    System.out.println("Enter how much money you will start with");
    System.out.println("Starting cash must be whole number");
    int money=0;

    do {
        try {
            System.out.println("Enter how much money you will start with: ");
            money = Integer.parseInt(input_var.next());

        }
        catch(NumberFormatException e) {
            System.out.println("Sry, not a valid integer");
            input_var.next();
        }

        input_var.next();

    } while (money<=0);

Upvotes: 1

Nikolay Bonev
Nikolay Bonev

Reputation: 151

Remove the input_var.nextInt(); from the try statement and input_var.nextLine();. There has to be an import like this import java.util.* or import java..util.InputMismatchException.

Which IDE are you using?

Upvotes: 1

Umar Farooq
Umar Farooq

Reputation: 331

You can use hasNextInt() within a while loop. While the next input is not an integer, display the "not a valid integer" message and get the next input. When it is an integer, the while loop will break and you can do what you need to do.

Upvotes: 1

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