CODEWITHSUNDEEP
javajava.util.scanner
starcorn
starcorn

Reputation: 8551

Scanner, useDelimiter

I encounter some problem when using useDelimiter from the Scanner class.

Scanner sc = new Scanner(System.in).useDelimiter("-");
while(sc.hasNext())
{
    System.out.println(sc.next());
}

if I have this input

A-B-C

the output will be

A B

and wait until I type in another "-" for it to print out the last character

However if I instead of having user input data, and insert a String to the Scanner instead the code will work. What's the reason for it, and how do I fix it? I don't want to use StringTokenzier

Upvotes: 2

Views: 19014

Answers (3)

user7598800
user7598800

Reputation:

How I would do it follows:

Scanner stdin = new Scanner(System.in);
String s = stdin.nextLine();
String[] splitInput = s.split("-", -1);

You will now have an array of Strings that contain the data between all of the -s.

Upvotes: 0

jjnguy
jjnguy

Reputation: 138922

If the Scanner didn't wait for you to enter another - then it would erroneously assume that you were done typing input.

What I mean is, the Scanner must wait for you to enter a - because it has no way to know the length of the next input.

So, if a user wanted to type A-B-CDE and you stopped to take a sip of coffee at C, it woud not get the correct input. (You expect [ A, B, CDE ] but it would get [ A, B, C ])

When you pass it in a full String, Scanner knows where the end of the input is, and doesn't need to wait for another delimiter.

How I would do it follows:

Scanner stdin = new Scanner(System.in);
String input = stdin.nextLine();
String[] splitInput = input.split("-", -1);

You will now have an array of Strings that contain the data between all of the -s.

Here is a link to the String.split() documentation for your reading pleasure.

Upvotes: 4

stacker
stacker

Reputation: 68972

You could use an alternative delimiter string useDelimiter( "-|\n" ); It works with a String argument as well as by reading from System.in. In case of System.in this requires you to press enter at the end of the line.

Upvotes: 3

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