Adding 2 binary numbers as string, C++

Question

I have a problem with adding 2 binary numbers. I want to do this as a string, so if they're different lengths I concatenate '0' to the beginning of shorter string. First of all, I don't know why, but I need to substrate then add '0' (without it, it wasn't working at all).

#include <iostream>
using namespace std;


string add( string no1, string no2 );


int equalizer(string no1, string no2)
{
int len1 = no1.length();
int len2 = no2.length();
if (len1 < len2)
{
    for (int i = 0 ; i < len2 - len1 ; i++)
    {

        no1 = '0' + no1;
    }
    return len2;
 }
 else if (len1 >= len2)
 {
    for (int i = 0 ; i < len1 - len2 ; i++)
    {

        no2 = '0' + no2;
    }
    return len1; // If len1 >= len2
 }

 }


 string add( string no1, string no2 )
 {
  string result="";


  int length = equalizer(no1, no2);

 int carry = 0;


 for (int i = length-1 ; i >= 0 ; i--)
 {
    int bit1 = no1.at(i) - '0';
    int bit2 = no2.at(i) - '0';

    // boolean expression for sum of 3 bits
    int sum = (bit1 ^ bit2 ^ carry)+'0';

    result = (char)sum + result;

    // boolean expression for 3-bit addition
    carry = (bit1 & bit2) | (bit2 & carry) | (bit1 & carry);
  } 

  // if overflow, then add a leading 1
  if (carry)
 {


    result = '1' + result;
  }

  return result;
  }
 bool check(string no1)
 {
 for(int i =0; i<no1.length(); i++)
 {
    if(no1.at(i)!=0 || no1.at(i)!=1)
    {
        cout << "not biniary! should contain only '0' and '1' "<< endl;
        return false;
    }
    else
    {
        return true;
    }
   }
  }
 int main()
 {
 string no1;
 string no2;
 cout << "Welcome to program that add 2 biniary numbers!" << endl;
 cout <<"Give first number " <<endl;
 cin >> no1;
 if(check(no1)==true)
 {





    cout <<"Give 2nd number" << endl;
    cin >> no2;
    check(no2);
    cout << "Numbers are proper!" << endl;
    add(no1,no2);
 }
 else
 {
    cout <<"End of program."<<endl;
 }
 return 0;
 }

W.F. · Accepted Answer · 2015-03-05 19:55:40Z

1

Instead of adding trailing zeros I would reverse the string at the beggining: std::reverse(s1.begin(), s1.end()). Then I would add each digit from the beggining. res[i] = s1[i] - s2[i] + '0'; Remember to initialize res string and then reverse it back again! Good luck!

answered Mar 5, 2015 at 19:55

W.F.

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2 Comments

batinex Over a year ago

I have changed it a little, now i've solved problem with adding biniary numbers but instead i ve got problems with expections. pastebin.com/PGLKRAS8 TBH i don't know what is wrong with expections. I know i know them less than little but i don't understand why e.g. 2 throw expection but 123 not?

W.F. Over a year ago

try to use gdb --args ./yourexecname then type run to access the backtrace type bt Of course to debug you need to add compiler options -g -O0 Good luck!

Satyam · Accepted Answer · 2017-06-17 06:48:15Z

0

use this method to add 0s in the beginning:-

considering that your str1 goes out of scope and str2 remains and you want to add str1 before str2

char *temp = strdup(str2);
strcpy(str2, str1); 
strcat(str2, temp);  
free(temp);

answered Jun 17, 2017 at 6:48

Satyam

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Comments

rashedcs · Accepted Answer · 2018-03-09 12:08:59Z

I think the following code can be helpful.

        int getBit(string s, int index)
        {
             if(index >= 0)   return (s[index] - '0');
             else             return 0;
        }

        string addBinary(string a, string b) 
        {
            if(a.size() > b.size())        while(a.size() > b.size()) b = "0" + b;
            else if(b.size() > a.size())   while(b.size() > a.size()) a = "0" + a;


            int l = max(a.size()-1, b.size() - 1);

            string result = ""; 
            int s=0;        

            while(l>=0 || s==1)
            {
                s += getBit(a, l) + getBit(b, l) ;
                result = char(s % 2 + '0') + result;
                s /= 2;
                l--;
            }
            return result;
        }

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