CODEWITHSUNDEEP
java
Anurag
Anurag

Reputation: 21

A program Binary to Decimal giving error in Java code

I wrote a program to convert a number From Binary to Decimal and it is giving wrong output if the input is 0011. For 0011 answer should be 3 but it is giving 9, otherwise it is correct for other input.

Code:

public class BinaryToDecimal {
static int testcase1=1001;
public static void main(String[] args) {
    BinaryToDecimal test = new BinaryToDecimal();
    int result = test.convertBinaryToDecimal(testcase1);
    System.out.println(result);
}   
//write your code here
public int convertBinaryToDecimal(int binary) {
    int powerOfTwo=1,decimal=0;
    while(binary>0)
    {
        decimal+=(binary%10)*powerOfTwo;
        binary/=10;
        powerOfTwo*=2;
    }
    return decimal;
}
}

Upvotes: 1

Views: 178

Answers (4)

Sandeep
Sandeep

Reputation: 24

Or Try This One Also 

import java.lang.*;
import java.io.*;

public class BinaryToDecimal{
 public static void main(String[] args) throws IOException{
BufferedReader bf= new BufferedReader(new InputStreamReader(System.in));
  System.out.print("Enter the Binary value: ");
  String str = bf.readLine();
  long num = Long.parseLong(str);
  long rem;
  while(num > 0){
  rem = num % 10;
  num = num / 10;
  if(rem != 0 && rem != 1){
  System.out.println("This is not a binary number.");
  System.out.println("Please try once again.");
  System.exit(0);
  }
  }
  int i= Integer.parseInt(str,2);
  System.out.println("Decimal:="+ i);
  }
        }

Upvotes: 0

Sandeep
Sandeep

Reputation: 24

Try it

public class Binary2Decimal
{
public static void main(String arg[])
{
    int sum=0,len,e, d, c;
    int a[]=new int[arg.length];

    for(int i=0; i<=(arg.length-1); i++)
    {
        a[i]=Integer.parseInt(arg[i]);        // Enter every digits i.e 0 or 1, with space 
    }
    len=a.length;
    d=len;
    for(int j=0; j<d; j++)
    {
        len--;
        e=pow(2,len);
        System.out.println(e);
        c=a[j]*e;
        sum=sum+c;
    }
    System.out.println("sum is " +sum);
}

static int pow(int c, int d)
{       
            int n=1;
    for(int i=0;i<d;i++)
    {
      n=c*n;
    }
    return n;
}
 }

Upvotes: 0

Sotirios Delimanolis
Sotirios Delimanolis

Reputation: 279970

An integer literal that starts with a 0 is considered to be an octal, ie. base 8.

0011

is

(0 * 8^3) + (0 * 8^2) + (1 * 8^1) + (1 * 8^0)

which is equal to 9.

0111

would be 

(0 * 8^3) + (1 * 8^2) + (1 * 8^1) + (1 * 8^0)

equal to 73.

You're looking for 0b0011 which is the integer literal representation of a binary number.

Upvotes: 5

kai
kai

Reputation: 6887

Do it with a String like this:

public static void main(String[] args) throws IOException 
{  
    boolean correctInput = true;
    BufferedReader m_bufRead = new BufferedReader(new InputStreamReader(System.in));
    System.out.println("Bitte geben sie eine Dualzahl ein:");
    String input = m_bufRead.readLine().trim();
    for(int i = 0; i < input.length(); i++) {
        if(input.charAt(i)!='0' && input.charAt(i)!='1') {
            correctInput = false;
        }
    }
    if(correctInput) {
        long dezimal = 0;
        for(int i = 0; i < input.length(); i++) {
            dezimal += Character.getNumericValue(input.charAt(i)) * Math.pow(2, input.length()-(i+1));
        }
        System.out.println("\nDezimalwert:\n" + dezimal);
    }
    else {
        System.out.println("Ihre Eingabe kann nicht umgewandelt werden");
    }
}

Upvotes: 1

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