AP CSA 2.10 Debugging: The Empty Substring

Unit 2: Selection and Iteration · Lesson 2.10 · Debugging

The Empty Substring

Implementing String Algorithms. This is not a blank editor: it is someone else's attempt. Find what is wrong, fix it, and submit to be graded against hidden test cases.

Back to the Lesson 2.10 page All Unit 2 lessons

Why debugging is its own skill

substring takes a start and a stop, and the stop is not included. Passing the same number twice asks for the characters from position i up to but not including position i, which is nothing at all. The program prints blank lines and no error.

What is wrong with it

The program should print each character of the word on its own line. It prints blank lines instead: substring is called with the same index for both arguments, which always produces an empty String. Fix the second argument.
The search should print the index where the target letter first appears, or -1 when it does not appear. The code adds 1 to the result unconditionally, which turns a not-found -1 into 0 and reports a real match at the wrong index. Remove the adjustment.
The length line is correct.

The program reads

A word, then a single character to search for.

The program should print

The length of the word, then each character on its own line, then the index of the first occurrence of the search character or -1.

Worked examples

These show what the FIXED program should print. There are more cases you cannot see, and they use different values, so patching around just these numbers will fail.

Example 1 input

cat x

Example 1 output

3
c
a
t
-1

Example 2 input

hello l

Example 2 output

5
h
e
l
l
o
2

Your answer

Main.java

import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner input = new Scanner(System.in);
    String word = input.next();
    String target = input.next();

    System.out.println(word.length());

    for (int i = 0; i < word.length(); i++) {
      System.out.println(word.substring(i, i));
    }

    System.out.println(word.indexOf(target) + 1);
  }
}

Input for the Run button

Run sends whatever is in the code box below, bugs and all, so you can see the crash or the wrong answer for yourself before you fix anything.

Stuck?

Hint 1

substring(a, b) returns the characters from index a up to but NOT including index b, so its length is b - a. When a and b are the same the length is zero. To get exactly one character starting at i, the stop has to be i + 1.

Hint 2

indexOf returns -1 to mean not found, and -1 is a signal, not a position. Adding 1 to it produces 0, which is a perfectly valid index and reads as a match at the very front of the word. Never do arithmetic on a sentinel value.

Hint 3

Check the not-found case deliberately. Searching a word for a letter it does not contain should print -1, not 0.

Where to go next

Rework the Lesson 2.10 page Try the from-scratch version of this lesson

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