CODEWITHSUNDEEP
javaclasssum
Bulba
Bulba

Reputation: 5

I'm having problems understanding how this Java works with the arguments I'm feeding it

The issue I'm having is understanding this code when running the command: java Driver 1000000 its returning: sum(1000000) = 1784293664

and no matter how long i try to look at it i cant understand why and how the code is doing this im just wondering if anyone can provide any help in understanding what this code actually does to the number?

class Sum
{
  private int sum;

  public int get() {
    return sum;
  }

  public void set(int sum) {
    this.sum = sum;
  }
}

class Summation implements Runnable
{
  private int upper;
  private Sum sumValue;

  public Summation(int upper, Sum sumValue) {
    this.upper = upper;
    this.sumValue = sumValue;
  }

  public void run() {
    int sum = 0;

    for (int i = 0; i <= upper; i++)
      sum += i;

    sumValue.set(sum);
  }
}

public class Driver
{
  public static void main(String[] args) {
    if (args.length != 1) {
      System.err.println("Usage Driver <integer>");
      System.exit(0);
    }

    if (Integer.parseInt(args[0]) < 0) {
      System.err.println(args[0] + " must be >= 0");
      System.exit(0);
    }

    // Create the shared object
    Sum sumObject = new Sum();
    int upper = Integer.parseInt(args[0]);

    Thread worker = new Thread(new Summation(upper, sumObject));
    worker.start();

    try {
      worker.join();
    } catch (InterruptedException ie) { }
    System.out.println("sum(" + upper + ") = " + sumObject.get());
  }
}

Thanks in advance

Andrew

Upvotes: 0

Views: 290

Answers (1)

rgettman
rgettman

Reputation: 178263

Summing the numbers 1 through 1 million:

(1 + 1000000) * 1000000 / 2 = 500000500000

This causes overflow in the int you're using to hold the sum. The result is:

500000500000 (mod 2^32) = 1784293664

Use a long to store the sum; it has a maximum value of 9223372036854775807 and can hold the sum.

Upvotes: 6

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