binary results are reversed

[STR9GNh0]

hi, i am trying to perform a binary conversion when i am stuck at a point where from Decimal i convert it to Binary the results will be in reversed. Is there a way to reverse the result to make it right?

my code is as follows:

int main ()
{
   int n = 0, number = 0;
 
   cout << "Enter a decimal number";
   cin >> number >> n;

int a;

if (number > 0)
{
   cout << "Binary representation: ";

   while (number > 0)
   {
      a = number % 2;	   	   	   	   
      number = number / 2;
      cout << a;
   }

   cout << endl;
}

return 0;
}

but my results eg: decimal i key in as: 589
results came out as: 1011001001 (its in reverse)
answer should be: 1001001101 instead.

i know by using a counter i could reverse it by recalculate the length of the result but i have no idea on how to do it and where to place it.

any help would be appreciated.

[jsmith]

build a std::string with the result instead of cout'ing it directly in the while loop.
The std::string will then be reversed.

Use std::reverse() to reverse it so it is in the correct order.

[STR9GNh0]

i cant use reverse. is there any other method?

oh and i forgot to add-in, string isn't allowed here as well.

[helios]

Then you'll just have to keep the results in an array and then read the array backwards for printing.

[STR9GNh0]

thanks for the help but how?

2. no alternative method such as using ++ to do it?

[helios]

The only other way would be to extract the bits from the most significant end, instead of the least significant end like you're doing, but that's not easier.

[STR9GNh0]

how should i go about??

[jsmith]

i & ( 1 << N )

gets you the Nth bit of i (0 or 1) where the LSB is bit #0.

[squeeb1985]

I think the best way would be to completely redo the way you go about it, finding the left-most digits first and outputting them in order. The following code will do it (although I'm guessing there is a more efficient way to write it):

#include<iostream>
#include<string>
#include<math.h>
using namespace std;

int main() //Start of main function
{

    //Variables
    int number,n=0,a,c;

    cout<<"Enter number to be converted to binary: ";
    cin>>number;//number to be converted to binary
    a=number; //placeholder for number variable
    n--;
    do 
    {
        n++;
        number=number/2;  //figure out value for n
    }while(number>0);
    number=a; //revert number back to original input
    do
    {
       c=pow(2,n); //c=2^n
       a=number/c;
       n--;
       cout<<a;
       number%=c;
    }while(n>=0);



    return 0;
}

[helios]

There's a much better way to get 2ⁿ than using pow(). Namely, what jsmith suggested.
1<<n == 2ⁿ