Decimal to Binary conversion using recursion

Question

#include<iostream>
#include<conio.h>
#include<math.h>
using namespace std;
long int d2b(int);
static long int binary=0;
static int i=0;
long int d2b(int num)
{
    if(num!=0)
    {
        binary=binary+pow(10,i)*(num%2);
        d2b(num/2);
        i++;
    }
    return(binary);
}

int main()
{
    int num;
    long int binary_ans=0;
    cout<<"Enter the number.";
    cin>>num;
    binary_ans=d2b(num);
    cout<<"Ans = "<<binary_ans;
    getch();
    return(0);
}

I am using Dev C++ compiler and this code doesnt seem to work. Can somebody please run this code on their compilers and give me a feedback. Also if the code seems incorrect to you, please tell me the reason why you think so.

What is the exact error that you are getting? Have you tried using a debugger? — Abhishek Bansal, Jul 07 '14 at 10:34
you're not using the result of `d2b(num/2);` anywhere and your `i` is always `0`... — Lanting, Jul 07 '14 at 10:35
@user1990169 I am not getting an error. Its just that the program wouldnt show the expected output. — Shreyash Saraf, Jul 07 '14 at 10:37
Firstly, conio.h doesn't work quite well with C++. and you are trying to print a binary representative, so better use a string return. Apart from that, try debugging yourself. — Shashish Chandra, Jul 07 '14 at 10:38
@Lanting I know I am not using the returned value anywhere. But i am updating the static binary variable each time. — Shreyash Saraf, Jul 07 '14 at 10:40
I'm uncertain what the d2b function is supposed to do. Even if you use conio correctly (as mentioned by Shashish Chandra) num will be binary already. — Evil Dog Pie, Jul 07 '14 at 10:42
@MikeofSST- Its supposed to update the static variable binary with every call. After its last call, i should have the binary output for any given decimal number. — Shreyash Saraf, Jul 07 '14 at 10:46
Thank you all for your help. An i++ statement was missing. I have updated the program and its working now. — Shreyash Saraf, Jul 07 '14 at 10:51

Lanting · Answer 1 · 2017-10-31T09:26:42.020

A correct implementation with neither static variables nor globals (both are just plain evil, try to avoid them at any cost) that gives you the desired output can be something as simple as

#include <iostream>

long int d2b(int x)
{
  return x ? (x%2 + 10*d2b(x/2)) : 0;
}

int main()
{
  std::cout << "enter a number:" << std::endl;
  int input;
  std::cin >> input;
  long int binary = d2b(input);
  std::cout << "ans = " << binary << std::endl;
}

Take note that, as we're using int to store our binary representation, the whole thing breaks for any input > 524287.

This of course assumes your integer is 64bit. The largest number you can write using only 0 and 1 is int x = 1111111111111111111, which when you interpret it as binary translates to 524287.

For 32bit and even 16bit integers this range is significantly lower.

Also have a look at: Are global variables bad?

+1. While `d2b()` isn't what I'd call *self-explanatory* but getting rid of globals is a huge improvement nonetheless. — TobiMcNamobi, Jul 07 '14 at 10:57

score 1 · Accepted Answer · answered Jul 07 '14 at 10:51

Obviously i++ must be done before the recursion takes place.

#include<iostream>
#include<conio.h>
#include<math.h>

using namespace std;

long int d2b(int);
static long int binary=0;
static int i=0;

long int d2b(int num)
{
    if(num!=0)
    {
        binary=binary+pow(10,i)*(num%2);
        i++;
        d2b(num/2);
    }
    return(binary);
}

int main()
{
    int num;
    long int binary_ans=0;
    cout<<"Enter the number.";
    cin>>num;
    binary_ans=d2b(num);
    cout<<"Ans = "<<binary_ans;
    _getch();
    return(0);
}

Get used to debugging your code!

Shashish Chandra · Answer 3 · 2014-07-07T11:01:42.243

As mentioned earlier, you can use it also in this manner:

#include<iostream>
#include<math.h>
using namespace std;
string d2b(int);
static string binary;
string d2b(int num)
{
    if(num!=0)
    {
        binary= to_string(num%2) + binary;
        d2b(num/2);
    }
    return(binary);
}

int main()
{
    int num;
    string binary_ans="";
    cout<<"Enter the number.";
    cin>>num;
    binary_ans=d2b(num);
    cout<<"Ans = "<<binary_ans;
    getch();
    return(0);
}

Decimal to Binary conversion using recursion

3 Answers3