nextLine() after nextInt() or nextDouble gives trouble

[Isky]

Hi,
I'm new to this Java world, and just can't figure out this problem. After I use nextInt or nextDouble, and try to use nextLine, it won't work unless I put nextLine twice. What's the problem there, and what would be the good way to handle this?

Just gave a code to illustrate a simple example I can understand.

import java.util.Scanner;

public class nextLineProblem {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		System.out.println("Type in something for line 1");
		String line1 = sc.nextLine();

		System.out.println("Type an integer for line 2");
		int line2 = sc.nextInt();
		
		System.out.println("Type in something for line 3");
		String line3 = sc.nextLine();
		
		System.out.println("Line1: " + line1 + "\n" + "Line2: " + line2 + "\n" + "Line3:" + line3);
		
	}

}

[jmaat7]

I had this problem before also.

this is what I came across:

http://www.robocommunity.com/blog/entry/12176/Java-Quick-Tip---Problems--with-the-Scanner-Class/

For my assignment a while back , I just used the .next() to read the next input. But my assignment only required a single word , not separated by spaces (so i could just use tokens).

looks like you're just going to have to add the extra line as far as I can tell.

[BestJewSinceJC]

The reason it gives you trouble is because when the user enters an integer then hits enter, two things have just been entered - the integer and a "newline" which is \n. The method you are calling, "nextInt", only reads in the integer, which leaves the newline in the input stream. But calling nextLine() does read in newlines, which is why you had to call nextLine() before your code would work. You could have also called next(), which would also have read in the newline.

[cgeier]

Just change the delimiter.

Scanner in = new Scanner(System.in);
in.useDelimiter("\n");

[Isky]

Thanks, now I understand what's going on now and how to deal with it when using the Scanner class.
Just left pondering if it's a bug or a feature :)

[Isky]

A new problem came to mind today.
It's easy to make a way around this problem if I know I'm going to have it, but what if it's unknown that I'll have a nextLine() after nextInt() , how would I be able to deal with the problem then?
For instance if there's a random function involved. Would I just have to scrap the nextLine() and go with something else?

[cgeier]

Check out the documentation for the scanner class.
http://java.sun.com/j2se/1.5.0/docs/api/java/util/Scanner.html

It will throw "NoSuchElementException" if no line was found. If you don't haven't learned about exception handling yet, I would use "hasNext".

int myInt = 0;
String myString = "";

Scanner in = new Scanner(System.in);
in.useDelimiter("\n");

if (in.hasNextInt()){
     //read in int value
     myInt = in.nextInt();
}//if
else if (in.hasNext()){
     //read in String value
     myString = in.next();
}//else if

[Isky]

No, haven't learned about exception handling yet, but I'll read up on that now. Thanks for pointing it out.

[morrisga2000]

THANKS SO MUCH FOR POSTING!!! I encountered the same problem and was getting so frustrated with the program not performing like I expected it to.
Sure enough, putting in nextLine() twice solved the problem.
I think that was is happening is that the nextInt and nextDouble are not properly terminating the allocated line of memory, so that when the nextLine() is run it is actually terminating the line that already has a value in it -- entered from the previous nextInt or nextDouble -- rather than taking in a new string readline value.