Ryan Harrison My blog, portfolio and technology related ramblings

Java - Common Pitfall when using Scanner

It’s very common for beginners in Java to use the many helpful methods of Scanner to obtain input from the user. They normally have some sort of task which involves asking the user for some String input, some numerical input and finally more String input - be this to do some processing with or just output back to the user in the simplest case. Most people will be happy using the nextLine() and nextInt() methods of Scanner to do so, however will then get confused when they don’t get the behaviour they expect. This is a great example of why you should know the specific behaviour of the library methods you call and not just assume that they will conform exactly to your needs by default.

As an example, say we get the task to ask the user for their name, storing it in a String variable, their age, storing it in an int variable, and finally their address, again storing it in a String variable. We may come up with this code to start off with:

  
import java.util.Scanner;

public class Sample {  
    public static void main(String[] args) {  
        Scanner scanner = new Scanner(System.in);  
        System.out.println("Enter name:");  
        String name = scanner.nextLine();

        System.out.println("Enter age:");  
        int age = scanner.nextInt();

        System.out.println("Enter address:");  
        String address = scanner.nextLine();

        System.out.println("—————");  
        System.out.println("Your name = " + name);  
        System.out.println("Your age = " + age);  
        System.out.println("Your address = " + address);  
    }  
}

Seems OK at a first glance, we prompt the user for input, use Scanner to retrieve it from the console and finally output what they entered. When we run this code however, it doesn’t function as expected:

  
C:\Users\Ryan\Java>java Sample  
Enter name  
Ryan  
Enter age  
20  
Enter address  
—————  
Your name = Ryan  
Your age = 20  
Your address =  

We get the name and age input successfully, but what happened to the address? We get prompted for the address yet have no opportunity to enter anything - resulting in a blank String for this field.

Although it may not seem like it, Scanner is actually behaving here exactly how it was designed to do so. The Scanner class is a multi-purpose tool, capable of not only parsing character input from the console, but also from other files and streams. In order to do so, it provides two methods: next() and nextLine(). The next() method takes a token from the current input stream up to the next delimiter (which by default is whitespace although this can be changed). It then returns this parsed token, but importantly leaves the delimiter in the stream. The nextLine() method on the other hand accepts a whole line up to and including the newline character. This time however, the delimiter gets consumed as well. Most of the other parsing methods in Scanner base themselves around these two base models.

Examples of those other parsing methods include convenient methods to accept input and convert it into another type (a call to nextInt() for example). Scanner does this by calling the next() method and then attempting to parse the output into the resulting type. This seems quite reasonable, but recall that we have called the next() method here and not nextLine(). This is where the problems occurs in our example above. The next() method leaves the delimiter in the stream, which means that when the user types a number, say “20” for their age and hits return, the Scanner gets a stream consisting of “20\n”. When the nextInt() call happens, the “20” is removed, and becomes a 20, and now the stream is “\n”. Now we call nextLine() again to retrieve the address. The Scanner does what it’s meant to and takes the the “\n”, discards the newline character (“\n”) and returns the remaining String - which in this case is “”. Thus we get the undesired behaviour.

The solution? Simply call nextLine() after we call nextInt() to consume the remaining newline character so that we get the chance to enter the address. Ideally however, you shouldn’t litter your input code with a load of new nextLine() calls to get the desired behaviour. Therefore, its probably best to implement your own utility class to take input in the manner you expect - leaving the Scanner in the correct state for further use afterwards.

This leaves the question as to why they decided to implement it in this way. As I said before Scanner is a very multi-purpose tool that should be able to accept a wide variety of input formats - not just one token on each line. A major use case of Scanner is to for example parse all the ints on one line, separated by whitespace, in order to populate some kind of Collection. If the nextLine() methods was used in nextInt() rather than next() we wouldn’t be able to do this and Scanner would be restrained in its use. This of course wouldn’t be a useful design decision and so we have to do this small bit of extra work to make it conform to our requirements.