Adding exception to java program

Question

I am having trouble adding an String exception to prevent my program from crashing when a string is entered instead of a int. I did look around and try out try{}catch{} but my program would still crash with a string. I'm looking to fix getInt().

import java.util.*;
public class Binary{

  public static void main(String[]args){
    Scanner in  = new Scanner(System.in);
    String a = "";
    boolean cont  = true;
    while(cont){
      printb(convert(getInt(in, "Enter a number: ")));
      System.out.println("Do you want to continue (yes or no)?");
      a = in.next();
      if(a.equals("yes"))
        cont = true;
      else{
        System.out.println("You answerd no. Have a nice day.");
        cont = false;
      }
    }


  }

  public static int getInt( Scanner console, String prompt){
    System.out.print(prompt);
    while(!console.hasNext()){
        try{
           console.next();
           System.out.println("Not an integer, try again.");
           System.out.print(prompt);
        }
        catch(Input MismatchException exception){
           System.out.print("Not an integer, try again.");
           System.out.print(prompt);
        }

    }
    return console.nextInt();
  }

  public static int[] convert(int decimal){

    int decimalCopy = decimal;
    int len = 0;
    while(decimal != 0){
      decimal/=2;
      len++;
    }
    decimal = decimalCopy;

    int[] b = new int[len];
    int index = 0;
    while(decimal !=0){
      if(decimal%2 != 0)
        b[index] = 1;
      else{
        b[index] = 0;
      }
    decimal/=2;
    index++;
    }

    return b;

  }

  public static void printb(int[] b){
    for(int i = b.length-1; i>=0; i--){
      System.out.print(b[i]);
    }
    System.out.println();
  }

} 

Answer 1

import java.util.*;
public class Binary{

  public static void main(String[]args){
    try {
    Scanner in  = new Scanner(System.in);
    String a = "";
    boolean cont  = true;
    while(cont){
      printb(convert(getInt(in, "Enter a number: ")));
      System.out.println("Do you want to continue (yes or no)?");
      a = in.next();
      if(a.equals("yes"))
        cont = true;
      else{
        System.out.println("You answerd no. Have a nice day.");
        cont = false;
      } }
    } catch(Exception e) {
      System.out.println("Invalid input");
    }


  }

  public static int getInt( Scanner console, String prompt){
    System.out.print(prompt);
    while(!console.hasNext()){
        console.next();
        System.out.println("Not an integer, try again.");
        System.out.print(prompt);
    }
    return console.nextInt();
  }

  public static int[] convert(int decimal){

    int decimalCopy = decimal;
    int len = 0;
    while(decimal != 0){
      decimal/=2;
      len++;
    }
    decimal = decimalCopy;

    int[] b = new int[len];
    int index = 0;
    while(decimal !=0){
      if(decimal%2 != 0)
        b[index] = 1;
      else{
        b[index] = 0;
      }
    decimal/=2;
    index++;
    }

    return b;

  }

  public static void printb(int[] b){
    for(int i = b.length-1; i>=0; i--){
      System.out.print(b[i]);
    }
    System.out.println();
  }

} `enter code here`

Answer 2

try/catch/finally is the way to handle this, but if you aren't familiar with exception handling, it can be tricky to figure out exactly what to do with them. And even when you do put them in the right place, you need to deal with the string input that hasn't been properly "cleaned up" so to speak, so cascades down to where a is assigned and ends the program (since everything that is not yes is no).

The key is to put the line(s) that might throw an exception inside the try block, then catch the Exception with some error handling, and then continue what needs to happen with finally. (You could omit the finally here, but I wanted to make sure you understand it, because it's important. finally comes after try/catch and anything within that code block will be executed in either scenario (unless you exit the program prematurely).)

This should do it:

while (cont) {

  // getInt() is the troublemaker, so we try it:
  // Notice I have changed 'number' to 'integer' here - this improves
  // UX by prompting the user for the data type the program expects:

  try {
    printb(convert(getInt(in, "Enter an integer: ")));
  }

  // we catch the Exception and name it. 'e' is a convention but you
  // could call it something else. Sometimes we will use it for info,
  // and in this case we don't really need it, but Java expects it
  // nonetheless.
  // We do our error handling here: (notice the call to in.next() -
  // this ensures that the string that was entered gets properly
  // handled and doesn't cascade down to the assignment of 'a') - if
  // this confuses you, try it without the in.next() and see what
  // happens!

  catch (Exception e) {
    System.out.println("\nPlease enter an integer.\n");
    in.next();
  }

  // Again, in this case, the finally isn't necessary, but it can be
  // very handy, so I'm using it for illustrative purposes:

  finally {
    System.out.println("Do you want to continue (yes or no)?");
    a = in.next();
    if (a.equals("yes")) {
      cont = true;
    } else {
      System.out.println("You answered no. Have a nice day.");
      cont = false;
    }
  }
}