Robert Martin in clean code: "I wrote the module in Listing 4-7...[as an] example of bad coding and commenting style. Kent Beck then refactored this code into a much more pleasant form in front of several dozen enthusiastic students. ...See how many different comment problems you can find...In Listing 4-8 you can see a refactored version"

Listing_4_7_GeneratePrime.java

/**
 * This class Generates prime numbers up to a user specified
 * maximum. The algorithm used is the Sieve of Eratosthenes.
 * <p>
 * Eratosthenes of Cyrene, b. c. 276 BC, Cyrene, Libya --
 * d. c. 194, Alexandria.  The first man to calculate the
 * circumference of the Earth. Also known for working on
 * calendars with leap years and ran the library at Alexandria.
 * <p>
 * The algorithm is quite simple. Given an array of integers
 * starting at 2. Cross out all multiples of 2. Find the next
 * uncrossed integer, and cross out all of its multiples.
 * Repeat untilyou have passed the square root of the maximum
 * value.
 *
 * @author Alphonse
 * @version 13 Feb 2002 atp
 */
import java.util.*;

public class GeneratePrimes
{
  /**
   * @param maxValue is the generation limit.
   */
  public static int[] generatePrimes(int maxValue)
  {
    if (maxValue >= 2) // the only valid case
    {
      // declarations
      int s = maxValue + 1; // size of array
      boolean[] f = new boolean[s];
      int i;
      // initialize array to true.
      for (i = 0; i < s; i++)
        f[i] = true;

      // get rid of known non-primes
      f[0] = f[1] = false;

      // sieve
      int j;
      for (i = 2; i < Math.sqrt(s) + 1; i++)
      {
        if (f[i]) // if i is uncrossed, cross its multiples.
        {
          for (j = 2 * i; j < s; j += i)
            f[j] = false; // multiple is not prime
        }
      }

      // how many primes are there?
      int count = 0;
      for (i = 0; i < s; i++)
      {
        if (f[i])
          count++; // bump count.
      }

      int[] primes = new int[count];

      // move the primes into the result
      for (i = 0, j = 0; i < s; i++)
      {
        if (f[i])             // if prime
          primes[j++] = i;
      }

      return primes; // return the primes
    }
    else // maxValue < 2
    return new int[0]; // return null array if bad input.
  }
}

Listing_4_8_PrimeGenerator.java

/**
 * This class Generates prime numbers up to a user specified
 * maximum.  The algorithm used is the Sieve of Eratosthenes.
 * Given an array of integers starting at 2:
 * Find the first uncrossed integer, and cross out all its
 * multiples.  Repeat until there are no more multiples
 * in the array.
 */

public class PrimeGenerator
{
  private static boolean[] crossedOut;
  private static int[] result;

  public static int[] generatePrimes(int maxValue)
  {
    if (maxValue < 2)
      return new int[0];
    else
    {
      uncrossIntegersUpTo(maxValue);
      crossOutMultiples();
      putUncrossedIntegersIntoResult();
      return result;
    }
  }

  private static void uncrossIntegersUpTo(int maxValue)
  {
    crossedOut = new boolean[maxValue + 1];
    for (int i = 2; i < crossedOut.length; i++)
      crossedOut[i] = false;
  }

  private static void crossOutMultiples()
  {
    int limit = determineIterationLimit();
    for (int i = 2; i <= limit; i++)
      if (notCrossed(i))
        crossOutMultiplesOf(i);
  }

  private static int determineIterationLimit()
  {
    // Every multiple in the array has a prime factor that
    // is less than or equal to the root of the array size,
    // so we don't have to cross out multiples of numbers
    // larger than that root.
    double iterationLimit = Math.sqrt(crossedOut.length);
    return (int) iterationLimit;
  }

  private static void crossOutMultiplesOf(int i)
  {
    for (int multiple = 2*i;
         multiple < crossedOut.length;
         multiple += i)
      crossedOut[multiple] = true;
  }

  private static boolean notCrossed(int i)
  {
    return crossedOut[i] == false;
  }

  private static void putUncrossedIntegersIntoResult()
  {
    result = new int[numberOfUncrossedIntegers()];
    for (int j = 0, i = 2; i < crossedOut.length; i++)
      if (notCrossed(i))
        result[j++] = i;
    }

  private static int numberOfUncrossedIntegers()
  {
    int count = 0;
    for (int i = 2; i < crossedOut.length; i++)
      if (notCrossed(i))
        count++;

    return count;
  }
}