Question

I am writing a program that accepts two ints within the program using nextInt(); and have it wrapped in a try catch block to stop bad inputs such as doubles or chars.

When multiple wrong inputs are entered the loop repeats that same number of times. I assume this is because my scan.next() has to loop around enough times to catch the bad inputs w/o error. Is there a way to know this number on the first run through to make a loop to run next in that many times?

In the output the
if(cont == 'N') System.out.print("\nPlease re-enter\n\t:"); will output and mirror the amount of times a mismatched input was written. That is, if I input 3 3.3 it will repeat one extra time, if input s 3.3 2.5 it will repeat three extra times.

I tried putting a loop around scan.next() to default it to ten times, but was overboard and I had to input an extra 8 characters before it started reading again. Maybe a while loop but what would its condition be, I tried while(scan.next() != null){} but that condition never stopped.

//input error checking
char cont = 'Y';
do{
  if(cont == 'N')
    System.out.print("\nPlease re-enter\n\t:");
  cont = 'Y';
  /* to stop the accidential typing of things other
   * than integers from being accepted
   */
  try{
    n1 = scan.nextInt();
    n2 = scan.nextInt();
  }catch(Exception e){
    cont = 'N'; //bad input repeat loop
    scan.next();//stops infinite loop by requesting Scanner try again
  }
} while(cont == 'N');//do loop while told N for continue

Answer 1

Not sure what you want your code to do. From reading what you have posted I assume you want the user to input 2 ints and if he/she doesn’t you want to prompt him/her to re-enter something until he/she inputs 2 ints.
If this is the case I would just add

scan = new Scanner(br.readLine());

after this if statement:

if(cont == 'N')
{System.out.print("\nPlease re-enter\n\t:");}

This will solve your looping issue