Thread: need some help with binary addition.

Kleid-0 have you tested your program with various inputs...I just compiled it with different values for b1 and b2 and it doesn't work.

I modified your code a bit to receive user input, and as you can see, it does not work. Even 1+1 or 0+0;

Code:

#include <stdio.h>
#include <stdlib.h>
#include <iostream>

using std::cout;
using std::cin;

// Allow boolean types


// Add binary numbers and return binary
int addBinary (int b1, int b2);

int main (void)
{
  bool notDone = true;
  // Binary numbers b1 and b2 get added together
  int b1, b2, b3;
  while( notDone )
  {
	cout << "\n1st: ";
	cin >> b1;
	
	if( b1 == 9 ) //input 9 for exit
		break;
	
	cout << "\n2nd: ";
	cin >> b2;

	

	b3 = addBinary (b1, b2);
	// Print out binary addition
	printf ("\n%d + %d = %d\n",b1, b2, b3);
  }
  return 0;
}

// Add binary numbers and return binary
int addBinary (
    int b1,     // The first binary number
    int b2)     // The second binary number
{
   // The binary addition answer, transforms into binary.
   int a;
   // Carrier for binary addition
   bool carrier = false;
   // Current digit step (starts at 1)
   int s = 2;

   // Add the numbers like decimal
   a = b1 + b2;

// Next Binary digit
NBD:
    // If the carrier is true, add one to the current placement
    if (carrier)
    {
      a += 1 * (s - (s / 2));
      carrier = false;
    }

    // If the carrier makes the postdigit a 3, we have trouble
    if (a % (s / 2 + s) < s / 2)
    {
      a += 1 * ((s * 10) - ((s * 10) / 2));
      a -= 1 * (s - (s / 2));
    }

    // If the addition yields a 2 * digit placement,
    // remove it and set carrier binary addition to true.
    if (a % s < (s / 2))
    {
      a -= 1 * (s - (s / 2));
      carrier = true;
    }

    // If our digit placement is high enough, exit
    if (s > a)
      return a;

    // Set the digit placement higher
    s *= 10;

    // Verify next placement (Next binary Digit)
    goto NBD;
}

You might want to edit your original post with the code stating that your solution does not work.

I dug and dug, and dug somemore...and finally, I found it! Not the exact solution to your problem but I knew I wrote something very similar a long time ago in my first c++ class. The assignemnt was a bit different, as I think one of the specs was that the binary number can only have a max of 4 digits (ie. 1011 is good, and 11101 is not) - this should be easily changed. Also, each input has to have leading zeros; so if you want to add 1 + 10, you should input 0001 + 0010 - I really don't remember why I did it this way - maybe becasue that the only way I knew how to

anywho I run the prog on some test input just now, and it seems to be working fine, but only with the rules I described above - no error handling whatsover (Hey it was my first c++ class, and one of the first assignments )

Code:

/*
 * Author: Jakub Bomba
 * Class: CS100, Prof. Bell
 * Date: 09.09.2002
 * Title: Binary Addition
 */

#include <iostream>
using std::cin;
using std::cout;
using std::endl;

/*
 * Function that actually does the binary addition
 * @param:	f - first binary number
 *			s - second binary number
 *			carry - flag for the carry
 */
int binaryAdd(int f, //first number
			  int s, //second number
			  int * carry) 
{

	int result=0;

	// exhaust all of the possible cases as described in the handout

	if( (f == 0) && (s == 0) ) 
	{
		result = ( *carry );
		*carry = 0;
	}

	else if( (f == 0) && (s == 1) ) 
	{ 		
		if( *carry==0 ) 
			result = 1;
		else 
		{
			result = 0;
			*carry = 1;
		}
	}
	else if( (f == 1) && (s == 0) ) 
	{ 
		if( *carry==0 ) 
		{
			result = 1;			
		} 
		else 
		{
			result = 0;
			*carry = 1;
		}
	} 
	else if( (f == 1) && (s == 1) ) 
	{ 
	
		if( *carry==0 ) 
		{
			result = 0;
			*carry = 1;
		} 
		else 
		{
			result = 1;
			*carry = 1;
		}
	}

	return result;
}


int main(void) { 

	int i;          // loop counter
	int carry=0;		// Initialize carry to 0 (IMPORTANT)

	int n1[4];	// MSB - index 0 LSB index 3 - MSB - Most significant bit  (leftmost)
	int n2[4];	// MSB - index 0 LSB index 3 - LSB - Least significant bit (rightmost)
	int result[5];	// MSB - index 0 LSB index 4

	char e_result[6]; // 5+null char
	char e_n1[5]; // 4+null char
	char e_n2[5]; // 4+null char

	// get data

	cout << "Enter first number: ";
	cin >> e_n1; 
	cout << "Enter second number: ";
	cin >> e_n2;

	// convert from char array to int array
	// It is useful to have data in int array when doing conversion into decimal 
	// value of a certain binary number which is not implemented in this program. 

	for(i = 0; i < 4; i++) 
	{
		n1[i] = (e_n1[i]-48);  
		n2[i] = (e_n2[i]-48);	
	}

	// initialize result array
	for( i = 0; i < 5; i++) 
		result[i] = 0;


	// do the addition	
	for( i = 3; i >= 0; i--) { // go from right to left (LSB to MSB)
		
		result[i+1] = binaryAdd(n1[i],n2[i],&carry);	// needs to add 1 to the counter
														// since LSB index for result 
														// is 1 more than LSB of operands		

		e_result[i+1] = result[i+1]+48; // For display purposes - convert to char array
	}

	result[0] = carry; // Add MSB of result
	e_result[0] = result[0] + 48; 
	e_result[5] = '\0'; // Put NULL Char

	

	cout << "   " << e_n1 << endl;
	cout << "  +" << e_n2 << endl;
	cout << "  -----" << endl;
	cout << "  " << e_result << endl << endl;

	return 0;

}

thanks for another tip and the boost, i was right when i said the solution uses logical operators but i was getting myself lost in the many if statements.