# Sum of two numbers without using arithmetic operators

Ex:
int a=10;
int b=10;
int sum=a+b;
without using "+" operator calculate sum

### Editorial / Best Answer

jintojos

• Member Since May-2008 | Jul 17th, 2008

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

Showing Answers 1 - 19 of 19 Answers

#### jintojos Profile Answers by jintojos

• Jul 17th, 2008

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

#### veeranarayana Profile Answers by veeranarayana

• Jul 17th, 2008

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;
}

#### ceren6 Profile Answers by ceren6

• Jul 18th, 2008

#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;
}

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#### ceren6 Profile Answers by ceren6

• Jul 21st, 2008

Operator Name Syntax Overloadable Included in C Arithmetic operators Unary Plus +a Yes Yes Addition (Sum) a + b Yes Yes Prefix Increment ++a Yes Yes Postfix Increment a++ Yes Yes Assignment by Addition a += b Yes Yes Unary Minus (Negation) -a Yes Yes Subtraction (Difference) a - b Yes Yes Prefix Decrement --a Yes Yes Postfix Decrement a-- Yes Yes Assignment by Subtraction a -= b Yes Yes Multiplication (Product) a * b Yes Yes Assignment by Multiplication a *= b Yes Yes Division (Quotient) a / b Yes Yes Assignment by Division a /= b Yes Yes Modulus (Remainder) a % b Yes Yes Assignment by Modulus a %= b Yes Yes

#### Saurabh Sharma Profile Answers by Saurabh Sharma

• Aug 14th, 2008

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

#### sriranga.ch Profile Answers by sriranga.ch

• Aug 22nd, 2008

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

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#### shiva chitta Profile Answers by shiva chitta

• Aug 25th, 2008

#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...........

#### rashmi.mohanty Profile Answers by rashmi.mohanty

• Sep 9th, 2008

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

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#### npatwardhan Profile Answers by npatwardhan

• Nov 2nd, 2008

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

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#### sarav957 Profile Answers by sarav957

• Jul 10th, 2009

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

#### ratnapaul Profile Answers by ratnapaul

• Jul 18th, 2009

`Codemain()<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 />}`

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#### luckypavan Profile Answers by luckypavan

• Oct 1st, 2010

```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;
}```

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#### indravardhans Profile Answers by indravardhans

• Jan 15th, 2011

```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;
}```

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#### rome2all Profile Answers by rome2all

• Aug 16th, 2011

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

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#### sravan

• Aug 20th, 2011

```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;
}
}```

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#### mukesh kumar

• Mar 25th, 2012

```Codeint main(){
int a,b;
printf("Enter the two numbers:
");

scanf("%d",&a);
scanf("%d",&b);
printf("Sum is: %d",add(a,b));
} ```

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#### mahamad

• Jun 14th, 2012

```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;
}```

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#### And

• Aug 28th, 2012

For two positive numbers:

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

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

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#### mukesh

• Aug 10th, 2015

Simple

```Code
void main()
{
int a=5,b=6,c;
c=a-(-b);
printf("%d",c)

}```

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