Decimal to Binary Conversion program

[realin]

hi guys

i am trying to make a program to convert decimal number into binary .. i am able to do that, but the number comes inverted.. like 1101 comes like 1011

now how do i swap it off..

here is teh code

import java.io.DataInputStream;
class dec2b{
 
public static void main(String a[]){
 
    DataInputStream inp=new DataInputStream(System.in);
    int nNum=0,nTemp=0,nSum=0,nBin=0,i=0;
 
 
    System.out.println("Enter a Number to Convert into Binary");
 
    try{nNum=Integer.parseInt(inp.readLine());}
    catch(Exception e){}
 
    while(nNum!=0)
    {
 
        nTemp=nNum;
 
        if(nTemp%2==1)
        nBin=1;
        if(nTemp%2==0)
        nBin=0;
 
        nSum=nSum*10+nBin;
        nNum=nNum/2;
 
    }
    System.out.println(nSum);
}}
 

please help me
thanks

[JosAH]

hi guys

i am trying to make a program to convert decimal number into binary .. i am able to do that, but the number comes inverted.. like 1101 comes like 1011

now how do i swap it off..

You don't; you should read the API documentation for the Integer class. It's all there.

kind regards,

Jos

[realin]

You don't; you should read the API documentation for the Integer class. It's all there.

kind regards,

Jos

hi jos,

I know there is a straight function integer(11,2) .. but i wanna make it thru logic..

cause that is my assignment.. thanks :)

[JosAH]

hi jos,

I know there is a straight function integer(11,2) .. but i wanna make it thru logic..

cause that is my assignment.. thanks :)

Recursion is your friend then (it avoids reversing the entire thing at the end).

Suppose you have a number 'n' less than 2: the result will be "0" or "1" then; otherwise
the result will be the binary string representation of the number n/2 appended with
the binary string representation of n%2. You can even avoid the division and
modulo by a bit of bit shifting (hint: >>> and <<).

kind regards,

Jos

[realin]

Recursion is your friend then (it avoids reversing the entire thing at the end).

Suppose you have a number 'n' less than 2: the result will be "0" or "1" then; otherwise
the result will be the binary string representation of the number n/2 appended with
the binary string representation of n%2. You can even avoid the division and
modulo by a bit of bit shifting (hint: >>> and <<).

kind regards,

Jos

thanks seems interesting.. will try :)

thanks thanks a lot

[r035198x]

hi jos,

I know there is a straight function integer(11,2) .. but i wanna make it thru logic..

cause that is my assignment.. thanks :)

Use BufferedReader or Scanner instead of DataInputStream for reading input.
To reverse an integer's digits, just make it a String, pass it to a StringBuilder and call the reverse method on it.

[r035198x]

Recursion is your friend then (it avoids reversing the entire thing at the end).

Suppose you have a number 'n' less than 2: the result will be "0" or "1" then; otherwise
the result will be the binary string representation of the number n/2 appended with
the binary string representation of n%2. You can even avoid the division and
modulo by a bit of bit shifting (hint: >>> and <<).

kind regards,

Jos

There you go again ...

[fern7teensexy08]

hello..can you tell me how to loop the decimal to binary conversion?

[JosAH]

hello..can you tell me how to loop the decimal to binary conversion?

What do you mean by "how to loop"? For all the rest: read this thread.

kind regards,

Jos

[jesz143]

/*
 * To change this template, choose Tools | Templates
 * and open the template in the editor.
 */
 
package decimalbinary;
import javax.swing.*;
/**
 *
 * @author Admin
 */
public class Main {
 
    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        // TODO code application logic here
 
        int x,i,q,z,temp=0;
        q=0; z=0;
       String a="";
x =Integer.parseInt(JOptionPane.showInputDialog("Please enter your number"));
 
    while (x!=0)
    {temp=x;
    x=temp/2;
     q=temp%2;
     a=q+a;
 
    }
    String msg="Output is "+a;
    JOptionPane.showMessageDialog(null, msg);
    }
 
}

[pbrockway2]

@jesz143
Accumulating the digits in a StringBuilder and then reversing it at the end (as suggested by r\d*x) would remove the need for all these string concatenations.

Is there a reason why this thread was woken from its well deserved rest?

[power885]

class Binary
{
    public static void main(String args[])
    {
    int result=0;
    int val=48,nbin=0,k=0;
    while(val!=0)
        {
 
            nbin=(val%2==0)?0:1;
                result=nbin*(int)Math.pow(10,k)+result;
                val/=2; 
        ++k;  
 
        }
    System.out.println(result);
    }
}
 

I hope this may work for you

[power885]

@JosAH
Hello JosAH,
I posted my solution for decimal to binary conversion.Since am newbie I need a suggestion to modify my solution

[jkmyoung]

pbrockway2: I like your solution. No need for recursion!
Modifying the original code a little:

StringBuffer sb = new StringBuffer();
  while(nNum!=0)  // code changes nNum
    { 
        if(nNum%2==1) 
          sb.append(1); 
        else //        if(nNum%2==0) 
          sb.append(0);
        nNum >>= 1;  // same as nNum /= 2; 
    }
sb.reverse();
 

[lokanath60]

This is the Program Which will convert a Decimal no to Binary fie ..

 
class Binary{
    public static void main(String[] args){
        int no=150;       //Input the No here
        int i=0;            
        int size=no/2;     //the no of 0 and 1 is no/2 approximetly;
        int a[]=new int[++size];    //creating an array of that size
        while(no>0){
 
            a[i++]=(no%2==0)?0:1;
        no /=2;
        }
 
        for(int j=i-1;j>=0;j--){
 
            System.out.print(a[j]);
 
        }
 
    }
}