AP CSA 2.8 Debugging: Zero Times Anything

Unit 2: Selection and Iteration · Lesson 2.8 · Debugging

Zero Times Anything

for Loops. 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 2.8 page All Unit 2 lessons

Why debugging is its own skill

An accumulator has to start at the identity for its operation. Starting a sum at zero is right and starting a product at zero is a guarantee that the answer is zero, forever, no matter how correct the rest of the loop is.

What is wrong with it

The factorial always prints 0. The product accumulator is initialized to 0, and zero multiplied by anything stays zero. Initialize it to the value that leaves a product unchanged.
The count of even numbers is one too high whenever n itself is even. The loop bound uses <= where the spec says to stop below n. Fix the bound.
The sum on the first line is correct. Notice that its accumulator starts at 0 and that this is right for a sum.

The program reads

A single integer n from 1 to 12.

The program should print

Three lines: the sum of 1 through n, then n factorial, then how many integers strictly below n are even.

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

Example 1 output

15
120
2

Example 2 input

Example 2 output

10
24
1

Your answer

Main.java

import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner input = new Scanner(System.in);
    int n = input.nextInt();

    int sum = 0;
    for (int i = 1; i <= n; i++) {
      sum = sum + i;
    }
    System.out.println(sum);

    int product = 0;
    for (int i = 1; i <= n; i++) {
      product = product * i;
    }
    System.out.println(product);

    int evens = 0;
    for (int i = 1; i <= n; i++) {
      if (i % 2 == 0) {
        evens++;
      }
    }
    System.out.println(evens);
  }
}

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

The identity for addition is 0 because adding 0 changes nothing. The identity for multiplication is 1 for the same reason. An accumulator always starts at the identity of the operation it accumulates.

Hint 2

The evens count says strictly below n. Test it with n = 4: the integers strictly below 4 are 1, 2 and 3, so the answer is 1. The starter includes 4 itself and reports 2.

Hint 3

Only two characters change in the whole file. If your fix is longer than that, re-read the two lines named in the task.

Where to go next

Rework the Lesson 2.8 page Try the from-scratch version of this lesson

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