Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

How does input.nextInt() work exactly?

Tags:

java

This is the program

public class bInputMismathcExceptionDemo {
public static void main(String[] args) {
    Scanner input = new Scanner(System.in);
    boolean continueInput = true;
    do {
        try {
            System.out.println("Enter an integer:");
            int num = input.nextInt();

            System.out.println("the number is " + num);
            continueInput = false;
        }
        catch (InputMismatchException ex) {
            System.out.println("Try again. (Incorrect input: an integer is required)");
        } 
        input.nextLine();
    }   
    while (continueInput);

}
}

I know nextInt() only read the integer not the "\n", but why should we need the input.nextLine() to read the "\n"? is it necessary?? because I think even without input.nextLine(), after it goes back to try {}, the input.nextInt() can still read the next integer I type, but in fact it is a infinite loop.

I still don't know the logic behind it, hope someone can help me.

like image 264
高亮节 Avatar asked Jan 26 '15 16:01

高亮节


People also ask

What happens when a non integer is read by a nextInt ()?

Because nextInt() will try to read the incoming input. It will see that this input is not an integer, and will throw the exception.

Does nextInt go to the next line?

nextInt() method only consumes the integer part and leaves the enter or newline character in the input buffer.

What does the keyboard nextInt do?

nextInt() Scans the next token of the input as an int. nextFloat() Scans the next token of the input as a float. nextLine() Advances this scanner past the current line and returns the input that was skipped. nextDouble() Scans the next token of the input as a double.


2 Answers

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.

like image 172
RealSkeptic Avatar answered Sep 24 '22 17:09

RealSkeptic


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.

like image 32
Maroun Avatar answered Sep 25 '22 17:09

Maroun