CODEWITHSUNDEEP
javadecimal
cdohara
cdohara

Reputation: 21

Making a binary to decimal converter

EDIT: Since there are claims that this is a duplicate question, I will state my stance on why it is not.

  1. The other question had posed a problem regarding the conversion of a string input to an integer (the answer deemed correct). This is irrelevant here, as the problem in my case is one of logic.

  2. The solutions on the other question cannot correct the problem this question presents (finding the syntactical "charAt" error).

  3. Frankly, the context of the other question is beyond the scope of my knowledge in Java.

I have successfully created a program that converts binary numbers to decimal numbers.

import java.util.Scanner;
public class ProblemThree // to convert binary to decimal
{
  public static void main (String[]args)
  {
    Scanner scan = new Scanner (System.in);
    System.out.print("Enter binary");
    int binary = scan.nextInt();
    int answer = 0;
    int length = String.valueOf(binary).length();
    int[] number = new int[length];
    int[] powerTwo = new int[length];
    
    // To list every digit separately //
    for (int n = 0; n <= length - 1; n++)
    {
      number[n] = (int) (binary / Math.pow(10, n)) % 10;
    }
    
    // To set the values in powerTwo to 2 //
    for (int n = 0; n <= length - 1; n++)
    {
      powerTwo[n] = 2;
    }
    
    // To update the values in powerTwo //
    for (int n = 0; n <= length - 1; n++)
    {
      if (n == 0)
        powerTwo[0] = 1;
      else if (n == 1)
        powerTwo[1] = 2;
      else
        powerTwo[n] *=  powerTwo[n-1];
    }
    
    // To add if current value is 1 //
    for (int n = 0; n <= length - 1; n++)
    {
      if (number[n] == 1)
        answer += powerTwo[n];
    }
    System.out.println(powerTwo[0]); // for testing
    System.out.println(number[0]); // for testing
    System.out.println("The decimal version of " + binary + " is " + answer + ".");
  }
}

Seeing how some of the steps were redundant, I tried to simplify the code.

import java.util.Scanner;
public class ProblemThreeTestFunction
{
  public static void main (String[]args)
  {
    Scanner scan = new Scanner (System.in);
    System.out.print("Enter binary");
    int binary = scan.nextInt();
    int answer = 0;
    int length = String.valueOf(binary).length();
    String number = String.valueOf(binary);
    
    for (int n = 0; n <= length - 1; n++)
    {
      if (number.charAt(n) == 1)
        answer += (int)Math.pow(2, n);
    }
    System.out.println("The decimal version of " + binary + " is " + answer + ".");
  }
}

I actually have no idea why the simplified code logic does not work. It uses the same concept as my original.

Upvotes: -3

Views: 221

Answers (3)

WasteD
WasteD

Reputation: 778

Why don't you just use:

int decimalValue = Integer.parseInt(binary, 2);

But if you use this you Need to use scan.nextLine() not scan.nextInt().

This will just convert your binary number to decimal using base 10 like seen here.

Upvotes: 3

Ted Hopp
Ted Hopp

Reputation: 234847

You need to test against the character '1':

if (number.charAt(n) == '1')

not the number 1 that you are currently doing:

if (number.charAt(n) == 1) // WRONG!

Also, since charAt counts from the left, and binary expansion counts from the right, you need to reverse the sense. Change

Math.pow(2, n)

to

Math.pow(2, length - 1 - n)

Upvotes: 1

Sagar U
Sagar U

Reputation: 96

You are comparing with digit 1, it should be character 1 i.e. '1'.

for (int n = 0; n <= length - 1; n++) {
  if (number.charAt(n) == '1') {
    answer += (int)Math.pow(2, length - n - 1);
  }
}

Also you need to correct the exponent part.

Upvotes: 1

Related Questions