adding two numbers

[vvidyadhara]

can i add two numbers without using arthemetic operators. can any one do it.

[Salem]

> can i add two numbers without using arthemetic operators
I've no idea what you can or cannot do, but since you're asking, my guess is you can't.

> can any one do it.
I can.
http://www.catb.org/~esr/faqs/smart-questions.html#prune

Hint

#include <stdio.h>
#include <conio.h>

unsigned int add(unsigned int a, unsigned int b)
{
  unsigned int c= 0;
  unsigned int r= 0;
  unsigned int t= ~0;
  for (t= ~0; t; t>>= 1)
  {r<<= 1;
    r|= (a^b^c)&1;
    c= ((a|b)&c|a&b)&1;
    a>>= 1;
    b>>= 1;}
  for (t=~0, c=~t; t; t>>= 1)
  {c<<= 1;
    c|= r&1;
    r>>= 1;}
  return c;}

void main ( void ) //so much greater than int main
{
  int t;
  t = add (7,3);
  printf("%d",t);
  getch(); //pause program
  system("pause"); //just incase getch don't work
}

[Aia]

In order for you to understand the answer to you question you need to know about XOR, AND and SHIFT bitwise operators.
Let's add 3 + 5:

integer1 =  3  = 0011b
integer2 =  5  = 0101b

    first operation        second operation      third operation
        0011                0011                 shift by one
        0101                0101                   
       ______              ______
 XOR    0110           AND  0001  ------>        <<1   0010

    Again...  
                   first operation   second operation   third operation
previous XOR        0110                0110            shift by one
previous <<1        0010                0010
                   ______              ______
               XOR  0100            AND 0010  ------>  <<1   0100

    Again...  
                   first operation  second operation    third operation
previous XOR        0100                0100            shift by one
previous <<1        0100                0100
                   ______              ______
                XOR 0000           AND  0100  ------>  <<1   1000

    Again...  
                   first operation   second operation   
previous XOR        0000               0000            
previous <<1        1000               1000
                    ______            ______
                XOR 1000          AND  0000  
When AND iguals 0 the result of XOR is iqual to the sum of the two integers

An example of that would be the following function:

#include <stdio.h>

unsigned long add(unsigned long integer1, unsigned long integer2)
{
    unsigned long xor, and, temp;
    
    and = integer1 & integer2; /* Obtain the carry bits */
    xor = integer1 ^ integer2; /* resulting bits */
    
    while(and != 0 ) /* stop when carry bits are gone */
    {
        and <<= 1; /* shifting the carry bits one space */
        temp = xor ^ and; /* hold the new xor result bits*/
        and &= xor; /* clear the previous carry bits and assign the new carry bits */
        xor = temp; /* resulting bits */
    }
    return xor; /* final result */
}

int main( void )
{
    int n1 = 5, n2 = 3;
    
    printf( "%d + %d = %d\n", n1, n2, add( n1, n2 ) );
    
    getchar();
    return 0;
}

[amitp49]

//No use '+' operator : can be implemented as full adder circuit diagram

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define MAX 10
int lastbit(int num)
{
    return (num & 0x01);
}

int sum_wo_op_use_circuit(int n1,int n2) //do not use '+' oprator for sum
{
    int carry=0,p=0,g=0,i=0,s=0,len=0,temp=0,res=0;
    int arr[MAX];

/*Assume : numbers are positive, else need to mul by -1*/ 

    while(n1 || n2 || carry)
    {
        printf("Lastbit of [%d]: %d\n",n1,lastbit(n1));
        printf("Lastbit of [%d]: %d\n",n2,lastbit(n2));
        p = lastbit(n1) ^ lastbit(n2);
        g = lastbit(n1) & lastbit(n2);

        arr[i++] = p ^ carry;
        temp = p & carry;

        carry = temp | g;   

        //shift one bit from both n1 & 2
        n1=n1>>1;
        n2=n2>>1;
    }

    len=i-1;
    //print the arr in reverse direction
    for(i=len;i>=0;i--)
    {
        printf("%d\t",arr[i]);
    }

    //make the number
    int mul = 1;
    for(i=0;i<=len;i++)
    {
        res += arr[i] * (mul<<i);
    }
    return res;
}

int main(void)
{
    int i=0;
    int num1=5,num2=7;
    i=sum_wo_op_use_circuit(num1,num2);
    printf("Final Sum ... [%d] + [%d] = %d\n",num1,num2,i);
    return 0;
}

[karthi.k.s]

#include<stdio.h>
#include<conio.h>
void main()
{
    int a,b;
    printf("Enter two numbers");
    scanf("%d%d",&a,&b);
    for(;b>=0;b--)/* ONLY RELATIONAL AND INCREMENT DECREMENT OPERATORS ARE USED HERE*/
    {
        a++;
    }
     printf("The sum is : %d",a);
 }