Sum of two numbers without using arithmetic operators

void main()
 {
          int a=10,b=20;
          while(b--) a++;
           printf("Sum is :%d",a);  
 }

int i,j;
if(i<j)
{
     for(i=0;i<=j;i++)
         j++;
    cout<<j;
}
else
{
   for(j=0;j<=i;j++)
       i++;
  cout<<i;
}

#include
int main()
{
unsigned int t = 0xffffffff;
unsigned int z = 1;
int a = 10, b = 10, sum = 0, r = 0;
while (t) {
int t1, t2;
t1 = a & 0x00000001;
t2 = b & 0x00000001;
if ((t1 ^ t2 ^ r) == 1) {
if (t1 == 1 && t2 == 1 && r == 1)
r = 1;
else
r = 0;
sum = sum | z;
} else {
if (t1 == 0 && t2 == 0 && r == 0)
r = 0;
else
r = 1;
}
a >>= 1;
b >>= 1;
z <<= 1;
t >>= 1;
}
printf("sum: %dn", sum);
return 0;
}

Use minus operator :-
int a=10,b=20,c;
c=a-(-b);

void main()
{
int a=10,b=10;
a=a<<1;
b=b<<1;
printf("Sum of a&b is %dn",(a&b));
}

#include add(int a,int b)
{
 if(!a) return b;
else return add((a&b)<<1,a^b);
}
void main()
{ int a=2, b=5,c; c= add(a,b);
cout<<c;
}


hope it 'll help you right...........

void main(){
    int a=5,b=7,i;
    for(i=1;i<=a;i++)
         b++;
    printf("sum value:%d",b);
}

Code
for (i = 0; i < a; i++) 
{
    count++;
}
for (j = 0; j < b; j++) 
{
    count++;
}
printf("%d", count);
 

Code
#include<stdio.h>
int main()
{
    int i = 1, j = 20, c;
    c = (i ^ j);
    printf("%d", c);
    return 0;
}
 

Code
main()<br />{<br />cout<<"Enter the numbers to be added";<br /> int a,b,x,y;<br /> cin>>a>>b;<br /> <br />do{<br /> x=a&b;<br /> y=a^b;<br /> x=a<<1;<br /> y=b;<br />}while(a);<br />cotu<<"The sum is:"<<y;<br />}

Code
#include<stdio.h>
main()
{
    int a, b, c, d, sum;
    printf("n Please enter a,b values:");
    scanf("%d %d", &a, &b);
    if (b > a) {
        c = b - a;
    } else {
        c = a - b;              // using minus operator 
    }
    d = square(b) - square(a);  // using square function and - minus operator
    sum = d / c;                // using division operator
    printf("The sum of a and b is:%d", sum);
    return 0;
}

Code
#include<stdio.h>
 
int result = 0X0;
int a, b, C = 0X0, aB, bB;
 
int main()
{
    int val = 0X1;
 
    scanf("%d%d", &a, &b);
    while (val) {
        aB = !!(a & val);
        bB = !!(b & val);
 
        if (aB ^ bB ^ C)
            result |= val;
 
        C = (aB & bB) | ((aB ^ bB) & C);
 
        val <<= 1;
    }
    printf("%d", result);
 
    return 0;
}

Code
class Test {
    public static void main(String ... z) {
        System.out.println(4 || 8);     //use logical or symbol u can add it 
}}

Code
         #include<iostream.h>
         void main()
 {
      int i=20,j=30,m,n;
                         m=i,n=j;
if(i<j)
{
          for(i=1;i<=m;i++)
                        j++;
         cout<<j;
}
else
{
        for(j=1;j<=n;j++)
                 i++;
  cout<<i;
}
 }

Code
int main(){
int a,b;
printf("Enter the two numbers: 
");
 
scanf("%d",&a);
scanf("%d",&b);
printf("Sum is: %d",add(a,b));
} 

Code
#include<stdio.h>
#include<conio.h>
main()
{
    int a=20,b=10,c;
    clrscr();
    c=a-~b-1;             //it will change the sign of operator truly magic
    printf("sum is %d",c);
    getch();
    return 0;
}

For two positive numbers:

int main(){
int a = 10;
int b = 10;

printf("%d",a ^ b | ((a & b)<< 1));
return 0;
}