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public class decimaltobinary {
	public static String printBinary(double num){
		StringBuilder binary = new StringBuilder();
		int intNum = (int)num;
		int count = 0;
		while (intNum != 0){
			binary.append(intNum %2);
			intNum = intNum /2;
		}
		binary.reverse(); // due to the conversion method, the binary is in backwards order
		binary.append(".");
		num = num - (int)num; // remove the numbers before the decimal, as they have already been converted.
		
		
		while (num > 0 || count < 4){
			if (count == 4){ break;}
			if (num <= 0) {
				binary.append(0);
			} else {
				double r = num * 2;
				if (r >= 1){ 
					binary.append(1);
					num = r - 1;
				} else {
					binary.append(0);
					num = r;
				}
			}
			count++;
			
		}
			
		
		
		return binary.toString();
	}
	public static void main(String[] args) {
		double numcheck = 42.3333333; // used solely for testing -- can be easily replaced with user input 
		System.out.println(String.valueOf(numcheck)+ "\t\t" +printBinary(numcheck));

	}

}

Here is my implementation using Java. You could always use Integer.toBinaryString(int) to convert the number before the decimal, but I believe that would be considered cheating in the form of a test/interview. Feel free to comment.

def dec2bin(number):
    ## integer part:
    int_number = int(number)
    int_bin_str = ''
    while int_number != 0:
        (int_number, remainder) = divmod(int_number, 2)
        int_bin_str = str(remainder) + int_bin_str

    ## fractional part
    frac_number = number - int(number)
    frac_bin_str = ''
    count = 0
    while( count < 4):
        frac_bin_str += str(int(2.0 * frac_number))
        frac_number  = 2*frac_number - int(2*frac_number)
        count += 1

    return int_bin_str+"."+frac_bin_str

## MAIN ##
print dec2bin(10)
print dec2bin(2.22)
print dec2bin(0.876)

#include <stdio.h>
#include <string>

void dectobin(const unsigned int& num)
{
    unsigned int n = num;
    std::string strout;

    while (0 != n) {
        if ((1 & n) == 1) {
            strout.insert(0, "1");
            printf("1\n");
        }
        else {
            strout.insert(0, "0");
            printf("0\n");
        }
        n = n >> 1;
    }

    printf("Decimal: %u Binary: %s\n", num, strout.data());
}

int main(int argc, char* argv[])
{
    int num = 120;
    dectobin((unsigned int)num);
    return 0;
}

How about this................

#include<iostream>
#include<string>
using namespace std;
int main(){
    unsigned int n=10;
    string s;
    while(n!=0){
          if((n&1)==1)
          s='1'+s;
          else
          s='0'+s;
          n=n>>1;
           }
           cout<<"s:"<<s<<endl;
    cin.get();
    }

Hi Everyone , converting numbers after decimal point is different from before decimal point. After decimal point we have to multiply each digit with 2 and is result is greater than or equal to 1, then its 1 else its 0.

#include<iostream>
#include<cmath>
using namespace std;
void binary(int n)         //prints binary representation of number before decimal point
{
	if(n>1)
	{
		binary(n/2);
	}
	cout<<n%2;
}

void bin_dec(double n)    // prints binary representation of number after decimal point
{
    for(int i=0;i<4;i++)	
	{
		n=n*2;
    	if(n>=1)
    	{
	    	cout<<1;
	    	n=n-floor(n);
	    	continue;
    	}
    	cout<<0;
    }
}

int main()
{
	double n,n2;
	int n1;
	cout<<"Enter the decimal number. \n";
	cin>>n;
	n1=floor(n);
	binary(n1);
	n2=n-n1;
	cout<<".";
	bin_dec(n2);
	return 0;
}