CODEWITHSUNDEEP
javabinarydecimal
user1701604
user1701604

Reputation: 1015

Binary to decimal conversion

I am a novice programmer trying to write a program that converts an entered Binary number into a decimal number. As far as I can tell, the math and code are right, and no compilation errors are returned, however the number that is output is NOT the correct decimal number. My code is as follows:

  String num;
  double result = 0;
  do {
  Scanner in = new Scanner(System.in);
  System.out.println("Please enter a binary number or enter 'quit' to quit: ");
  num = in.nextLine();
  int l = num.length();
  if (num.indexOf("0")==-1 || num.indexOf("1")==-1 ){
    System.out.println("You did not enter a binary number.");
  }

  for (int i = 0; i < l; i++)
{ 
  result = result + (num.charAt(i) * Math.pow(2, (l - i)));
}
System.out.println("The resulting decimal number is: " +result);
  } while (!num.equals("quit"));


  if (num.equals("quit")){
    System.out.println("You chose to exit the program.");
    return;
  }

Any Help you could give would be much appreciated. I tried making my question as clear as possible but if you have any questions I will try to answer the best I can. I have not been doing this long. All I need is for someone to look it over and hopefully find the error I made somewhere, thanks.

Upvotes: 0

Views: 2370

Answers (4)

Tulains C&#243;rdova
Tulains C&#243;rdova

Reputation: 2639

Integer.parseInt(string, base) parse string into an integer using "base" radix, if it cannot be converted, it raises an exception.

import java.util.Scanner;

public class Convertion {

    public static void main(String[] args) {
        String num;
        Scanner in = new Scanner(System.in);
        System.out.println("Please enter a binary number");
        num = in.nextLine();
        try{
              //this does the conversion
              System.out.println(Integer.parseInt(num, 2));
        } catch (NumberFormatException e){
              System.out.println("Number entered is not binary");
        }  
    }
}

Upvotes: 0

arshajii
arshajii

Reputation: 129587

Change

result = result + (num.charAt(i) * Math.pow(2, (l - i)));

to

result = result + ((num.charAt(i) - '0') * Math.pow(2, i));

or more compactly,

result += (num.charAt(i) - '0') * Math.pow(2, i);

Remember that the character '0' is not the same thing as the number 0 (same for '1' and 1); num.charAt(i) returns the character not the integer.

int a = '0';
int b = 0;
System.out.println(Math.pow(2, a));
System.out.println(Math.pow(2, b));

Output:

2.81474976710656E14
1.0

Big difference isn't there?

Upvotes: 2

Stefan Neubert
Stefan Neubert

Reputation: 1054

The function String.charAt(); does not return the number 0 or 1 which you can multiply with the bit but the character "id". You need to convert the String / char to a number.

String num;
  double result = 0;
  do {
  Scanner in = new Scanner(System.in);
  System.out.println("Please enter a binary number or enter 'quit' to quit: ");
  num = in.nextLine();
  int l = num.length();
  if (num.indexOf("0")==-1 || num.indexOf("1")==-1 ){
    System.out.println("You did not enter a binary number.");
  }

  for (int i = 0; i < l; i++)
{ 
  result = result + (Integer.parseInt(num.substring(i,i+1)) * Math.pow(2, (l - i)));
}
System.out.println("The resulting decimal number is: " +result);
  } while (!num.equals("quit"));


  if (num.equals("quit")){
    System.out.println("You chose to exit the program.");
    return;
  }

By the way: Why is a string that does not contain either a 0 or a 1 not a binary numer? Think of 1111 for example. I think you should better check for "neither 0 nor 1"

if (num.indexOf("0")==-1 && num.indexOf("1")==-1 ){
    System.out.println("You did not enter a binary number.");
  }

Upvotes: 1

Code-Apprentice
Code-Apprentice

Reputation: 83577

Note that num.charAt(i) gives the ASCII code for the character at position i. This is not the value you want. You need to convert each character digit to an int before you do any math with the value.

Upvotes: 0

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