Binary addition of 2 values represented as strings

Question

I have two strings:

string a = "00001"; /* which is decimal 1 I've converted with next string:
string a = Convert.ToString(2, 2).PadLeft(5, '0'); */
string b = "00010";

I want to perform binary addition between the two so the answer will be 00011 ( 3).

Homework it might be, but still quite an interesting one - it piqued my curiosity enough that I've got a Console app open and fiddling around with it right now! ;) — Rob
– Rob, Commented Feb 12, 2010 at 15:34
Actually, looking at your comments on the accepted answer I think you might be looking in the wrong direction. You're not really looking at addition I think, but at a bitwise AND. — Erik van Brakel
– Erik van Brakel, Commented Feb 17, 2010 at 15:53
@Erik i was looking in the wrong direction true but thanks to this answer i now made a new function that takes 2 integers and returns a string equivalent with the proceeding zero's , which is exactly what i needed. — Alfred
– Alfred, Commented Feb 17, 2010 at 17:05

Benoît Vidis · Accepted Answer · 2010-02-12 15:37:17Z

21

System.Convert should be able to do the work for you

int number_one = Convert.ToInt32(a, 2);
int number_two = Convert.ToInt32(b, 2);

return Convert.ToString(number_one + number_two, 2);

(you may have to tune the strings a bit)

answered Feb 12, 2010 at 15:37

Benoît Vidis

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3 Comments

Alfred Over a year ago

string a = "00001"; string b = "00011"; int num1 = Convert.ToInt32(a, 2); int num2 = Convert.ToInt32(b, 2); string ans = Convert.ToString(num1 + num2, 2); MessageBox.Show(ans); "thanks alot :) u have saved my project !!!!!!!!!!!!!!!!"

Alfred Over a year ago

i needed to caluclate attendance using this logic :) , as 00001 wld represent absent for 1st hour ( we haf hourly wise attendance) so if a student is absent for the next hr in the same day 00001 - absent first hr 00010 - absent second hr ------ 00011 - absent for 1st and second hr :D it works !! thank u

Rob Over a year ago

I have to say, I didn't know that particular override of Convert.ToInt32 exists!

IVlad · Accepted Answer · 2010-02-12 15:30:06Z

8

You do it just as you would do it on paper. Start from right and move left. if A[i] + B[i] + carry >= 2, carry remains 1 and you move on. Otherwise, write A[i] + B[i] + carry and set carry to 0.

a = "00001"; b = "00010";

carry = 0; a[4] + b[4] + carry = 1, write 1, set carry = 0: 00001

a[3] + b[3] + carry = 1, write 1, set carry = 0: 00011

And so on.

answered Feb 12, 2010 at 15:30

IVlad

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Comments

Ronnie Diaz · Accepted Answer · 2011-09-26 13:45:40Z

2

private static bool[] BinaryAdd(bool[] originalbits, long valuetoadd)
    {
        bool[] returnbits = new bool[originalbits.Length];

        for (long i = 0; i <= valuetoadd - 1; i++)
        {
            bool r = false; //r=0
            for (long j=originalbits.Length-1;j<=originalbits.Length;j--)
            {
                bool breakcond = false;
                bool o1 = originalbits[j];
                if (r == false)
                {
                    if (o1 == false) { o1 = true; breakcond = true; }//break
                    else if (o1 == true) { o1 = false; r = true; }
                }
                else
                {
                    if (o1 == false) { o1 = true; breakcond = true; }//break
                    else if (o1 == true) { o1 = false; r = true; }
                }

                originalbits[j] = o1;
                if (breakcond == true)
                {
                    break;
                }
            }

        }
        returnbits = originalbits;

        return returnbits;
    }

answered Sep 26, 2011 at 13:45

Ronnie Diaz

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1 Comment

titanofold Over a year ago

You should add some explanation to your example so novices can understand what it does.

ihsan güç · Accepted Answer · 2022-01-10 10:29:54Z

1

public static string AddBinary(string a, string b)
    {
        string result = "";
         
        int s = 0;
         
        int i = a.Length - 1, j = b.Length - 1;
        while (i >= 0 || j >= 0 || s == 1)
        { 
            s += ((i >= 0) ? a[i] - '0' : 0);
            s += ((j >= 0) ? b[j] - '0' : 0);
            result = (char)(s % 2 + '0') + result;

            s /= 2;
            i--; j--;
        }
        return result;
    }

Just it. Extra Link.

answered Jan 10, 2022 at 10:29

ihsan güç

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Comments

user229044 · Accepted Answer · 2012-10-10 16:39:05Z

0

Very easy -- write a lookup table for 'addition' of binary characters, don't forget to carry if necessary, and send me 50% of the credit you get for the work.

edited Oct 10, 2012 at 16:39

user229044♦

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answered Feb 12, 2010 at 15:30

High Performance Mark

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Comments

krazor · Accepted Answer · 2018-06-01 12:12:21Z

0

var sum = Convert.ToString(Convert.ToInt32("00010", 2) + Convert.ToInt32("00001", 2), 2).PadLeft(5, '0');

"00011"

PadLeft is not really needed, but if the strings had different length you would get a the same format. Example:

var sum = Convert.ToString(Convert.ToInt32("0000010", 2) + Convert.ToInt32("001", 2), 2).PadLeft(5, '0');

"00011"

edited Jun 1, 2018 at 12:12

answered Jun 1, 2018 at 12:03

krazor

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Comments

C. Ross · Accepted Answer · 2010-02-12 15:30:38Z

-1

I would recommend parsing the data to ints and then adding them, then outputing the result as binary.

answered Feb 12, 2010 at 15:30

C. Ross

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Comments

Jeffrey L. Roberts · Accepted Answer · 2023-08-20 18:36:40Z

-1

If a[4] + b[4] + carry = 3 then write 1, carry 1 

Else If a[4] + b[4] + carry = 2 then write 0, carry 1 

Else If a[4] + b[4] + carry = 1 then write 1, carry 0

answered Aug 20, 2023 at 18:36

Jeffrey L. Roberts

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