Why do i have to use nextLine() when there is an InputMismatchException

Question

I dont understand why do i have to use this line of code to avoid an infinite loop here...help appreciated.
import java.util.Scanner;
import java.util.InputMismatchException;
public class InputTest {
     public static void main(String [] args){
     Scanner in = new Scanner(System.in);
     int n = 0;
     while(true){
          try{
               System.out.println("Input a number:");
               n = in.nextInt();
               break;
          }catch(InputMismatchException exc){
               System.out.println("Error, non-valid input.");
               //in.nextLine(); <--If i use this line of code it doesn't get into an infinite loop
                         System.out.println("Integer read is: " + n);
}

Answer 1

Rix87 wrote:
When i run the program if i input something that triggers the exception, it infinitely loops through the question and the catch method block outputs, weirdest thing of all is that i can't even input again after the exception is thrown, but with nextLine() it avoids that, but i still dont understand why do i have to use nextLine().Exactly what I said.
You enter "A". The exception is thrown. The "A" is not consumed. It calls nextInt on "A" again... and again... and again...
If you do nextLine, the "A" and the newline are consumed.
Even on good input, I don't think nextInt consumes the EOL.