Scanner Input, Tokens and StringTokenizer (ISC Class 12 Computer Science)

Understand Scanner Input, Tokens and the StringTokenizer Class

Scanner Input, Tokens and the StringTokenizer Class: key facts

Scanner methods

• •nextInt() reads an int
• •nextDouble() reads a double
• •next() reads one token
• •nextLine() reads the rest of the line

Package java.util.

StringTokenizer methods

• •countTokens()
• •hasMoreTokens()
• •nextToken()
• •Constructors: (str), (str, delims), (str, delims, true)

countTokens() reduces as tokens are taken.

Input exceptions

InputMismatchException	Scanner got the wrong kind of token
IOException	stream input or output failure
NoSuchElementException	no token left

IOException is checked and must be caught or declared.

Delimiters

Default delimiters	space, tab, newline
Delimiter string	each character is a delimiter
Third argument true	delimiters are returned as tokens

Consecutive delimiters produce no empty token.

How to answer Scanner Input, Tokens and the StringTokenizer Class questions

1. 1Import java.util.* (and java.io.* if you use BufferedReader).
2. 2Create the Scanner; read numbers with nextInt or nextDouble and text with next or nextLine.
3. 3After a number and before a line of text, add a nextLine() call to consume the leftover newline.
4. 4For a sentence, read one line, build a StringTokenizer, store countTokens() first, then loop with hasMoreTokens().
5. 5Convert numeric tokens with Integer.parseInt or Double.parseDouble.
6. 6Wrap risky input in try-catch for InputMismatchException or NumberFormatException when the question asks.

Scanner Input, Tokens and the StringTokenizer Class in real life

The idea, met in an everyday situation. Answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Tricks to remember Scanner Input, Tokens and the StringTokenizer Class

Short memory aids. Say each one aloud once, then use it on a question the same day.

• next stops, nextLine eats the line

  next() stops at whitespace; nextLine() runs to the end of the line.

  Use it when: Choosing the right read method.

• Count first

  Take countTokens() before the loop, because it shrinks as tokens are used.

  Use it when: Counting words in a sentence.

• Each delimiter character counts

  In (str, ",;") the comma and the semicolon are two separate delimiters.

  Use it when: Tokenizer with several delimiters.

• IO is checked

  IOException must be caught or thrown; InputMismatchException need not be.

  Use it when: Exception questions.

Scanner Input, Tokens and the StringTokenizer Class: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

StringTokenizer st = new StringTokenizer("12,5;7", ",;");
int n = st.countTokens();
int sum = 0;
while (st.hasMoreTokens())
    sum += Integer.parseInt(st.nextToken());
System.out.println(n + " " + sum);

Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
String s = sc.nextLine();
System.out.println("[" + s + "]");

import java.util.*;
public class Words {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String line = sc.nextLine();
        StringTokenizer st = new StringTokenizer(line);
        int n = st.countTokens();
        String longest = "";
        while (st.hasMoreTokens()) {
            String w = st.nextToken();
            if (w.length() > longest.length())
                longest = w;
        }
        System.out.println(n + " " + longest);
    }
}

Where students lose marks in Scanner Input, Tokens and the StringTokenizer Class

The slips examiners see most often, and what to write instead.

• ✗Mistake: Counting tokens after the loop

  countTokens() returns the remaining tokens, which is 0 once the loop has finished.

  Wrong working

  while (st.hasMoreTokens()) { ... } n = st.countTokens();

  Right working

  n = st.countTokens(); while (st.hasMoreTokens()) { ... }

  ✓Do this: Store the count before using the tokens.

  Usually costs: 1 mark

• ✗Mistake: Forgetting to convert a token

  nextToken() returns a String, which cannot be added as a number.

  Wrong working

  sum += st.nextToken();

  Right working

  sum += Integer.parseInt(st.nextToken());

  ✓Do this: Use parseInt or parseDouble.

  Usually costs: 1 mark

• ✗Mistake: No import

  Scanner and StringTokenizer are not in java.lang, so the program does not compile.

  Wrong working

  Scanner sc = new Scanner(System.in); with no import

  Right working

  import java.util.*; before the class

  ✓Do this: Write the import line first.

  Usually costs: 1 mark

Practice questions on Scanner Input, Tokens and the StringTokenizer Class

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   Which StringTokenizer method returns the number of tokens still to be read?
   (a) length() (b) countTokens() (c) size() (d) tokens()

   Hint

   It is named for the tokens.

   Answer

   (b) countTokens()

2. Q2 · 2 marks · Short answerBoard level

   Distinguish between next() and nextLine() of the Scanner class.

   Hint

   Think of what stops each one.

   Answer

   next() returns the next token and stops at whitespace, so it cannot read a space inside the text. nextLine() returns everything up to the end of the line, including spaces, and consumes the newline.

3. Q3 · 3 marks · Short answerHarder board level

   How many tokens does StringTokenizer give for the String "a+b-c" with (i) the default delimiters, (ii) delimiters "+-", (iii) delimiters "+-" with returnDelims true? List the tokens in (iii).

   Hint

   There is no whitespace in the String.

   Answer

   (i) 1 token, the whole String, because it has no whitespace. (ii) 3 tokens: a, b, c. (iii) 5 tokens: a, +, b, -, c.

4. Q4 · 3 marks · Short answerHarder board level

   Write the output:

   StringTokenizer st = new StringTokenizer("one two three");
   String s = "";
   while (st.hasMoreTokens())
       s = st.nextToken().charAt(0) + s;
   System.out.println(s);

   Hint

   Each new character goes in front of the old text.

   Answer

   Tokens: one, two, three. s = "o"; then "t" + "o" = "to"; then "t" + "to" = "tto". Output: tto

5. Q5 · 4 marks · Case-basedHOTS / competency

   A program reads an integer age and then a full name using sc.nextInt() and sc.nextLine(). (i) State the fault in the output. (ii) Rewrite the three input lines so that both values are read correctly. (iii) Give one other way to avoid the fault.

   Hint

   The Enter key typed after the number.

   Answer

   (i) The name is empty because nextLine reads the leftover end of the age line. (ii) int age = sc.nextInt(); sc.nextLine(); String name = sc.nextLine(); (iii) Read the age with nextLine() and convert it using Integer.parseInt(), so no newline is left behind.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.What are the default delimiters of StringTokenizer?Show answer

   Space, tab, newline and similar whitespace characters.

2. 2.What does nextToken() return?Show answer

   The next token as a String.

3. 3.What is thrown when nextInt() meets the text abc?Show answer

   InputMismatchException.

4. 4.Is IOException checked?Show answer

   Yes, it must be caught or declared.

5. 5.What does nextLine() return after nextInt()?Show answer

   The empty remainder of the same line.

6. 6.In which package is StringTokenizer?Show answer

   java.util.

Scanner Input, Tokens and the StringTokenizer Class: frequently asked questions