How does input.nextInt() work exactly?

Question

This is the program

public class bInputMismathcExceptionDemo {
public static void main(String[] args) {
    Scanner input = new Scanner(System.in);
    boole         


        

Answer 1

but why should we need the input.nextLine() to read the "\n"? is it necessary??

Yes (actually it's very common to do that), otherwise how will you consume the remaining \n? If you don't want to use nextLine to consume the left \n, use a different scanner object (I don't recommend this):

Scanner input1 = new Scanner(System.in);
Scanner input2 = new Scanner(System.in);

input1.nextInt();
input2.nextLine();

or use nextLine to read the integer value and convert it to int later so you won't have to consume the new line character later.

Answer 2

The reason it is necessary here is because of what happens when the input fails.

For example, try removing the input.nextLine() part, run the program again, and when it asks for input, enter abc and press Return

The result will be an infinite loop. Why?

Because nextInt() will try to read the incoming input. It will see that this input is not an integer, and will throw the exception. However, the input is not cleared. It will still be abc in the buffer. So going back to the loop will cause it to try parsing the same abc over and over.

Using nextLine() will clear the buffer, so that the next input you read after an error is going to be the fresh input that's after the bad line you have entered.