## INDEX

### Write a c program to add two numbers without using addition operator

Add two numbers in c without using operator

How to add two numbers without using the plus operator in c

#include<stdio.h>

int main(){

int a,b;
int sum;

printf("Enter any two integers: ");
scanf("%d%d",&a,&b);

//sum = a - (-b);
sum = a - ~b -1;

printf("Sum of two integers: %d",sum);

return 0;
}

Sample output:

Enter any two integers: 5 10

Sum of two integers: 15

Algorithm:

In c ~ is 1's complement operator. This is equivalent to:
~a = -b + 1
So, a - ~b -1
= a-(-b + 1) + 1
= a + b – 1 + 1
= a + b

Unknown said...

sir (~) tiled use karne se two number add kyun hue...kya aap xplain kr sakte h...

Unknown said...

@mohit
a+b=a-(-b)
but we know -b=(~b+1)
so
a+b=a - (~b+1)
a+b= a - ~b - 1

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Unknown said...

here is a simple logic...adding a and b..say a=5 b=4 so a+b=5+4=9

int main(void)
{
int a=5,b=4;// you can input those nos from user...
while(b)
{
a++;
b--;
}
printf("%d",a);
}

Anonymous said...

still not clear with '~'...

Anonymous said...

~b = -(b+1)
so a-(~b)-1 = a-(-(b+1))-1
= a+b+1-1
= a+b

Anand said...

'~' is a type of bitwise operator.
This simply means one's complement.
Bitwise operators can only be operated upon ints & chars.
On taking the one's complement of a number, all the 1's are changed to 0's and vice-versa.
e.g: one's complement of 5 (0101) is -6 (1010).

Anand said...

This code doesn't use any arithmetic operators to add two numbers.

main()
{
int num1, num2;
scanf("%d %d",num1, num2);
}

{
if (y == 0)
return x;
else
return Add( x ^ y, (x & y) << 1);
}

Unknown said...

please write explanation also. . .

Mukilan said...

while (num2) // do until carry
{
int carry = num1 & num2 // if its 1 & 1 we get carry
num1 = num1 ^ num2; // add all 0 +1 = 1
num2 = carry << 1; // now we need to carry
}
return num1

Mukilan said...

while (num2) // do until carry
{
int carry = num1 & num2 // if its 1 & 1 we get carry
num1 = num1 ^ num2; // add all 0 +1 = 1
num2 = carry << 1; // now we need to carry
}
return num1

Mukilan said...

while (num2) // do until carry
{
int carry = num1 & num2 // if its 1 & 1 we get carry
num1 = num1 ^ num2; // add all 0 +1 = 1
num2 = carry << 1; // now we need to carry
}
return num1

Anonymous said...

BH JUYHC

Anonymous said...

i cant understand

Anonymous said...

plz also share the problem for multiplication with out using * operator!!!

Unknown said...

Here is a slight modification that adds subtraction feature as well:

/* Adds two signed integers. Does a subtract when cin = 1 */
int adder(int a, int b, int cin)
{
if (cin) b = ~b;
int carry = (a & b) << 1 | cin;
a ^= b;
b = carry;

while (b) {
carry = (a & b) << 1;
a ^= b;
b = carry;
}

return a;
}

Anonymous said...

you can as well use ^ operator.
a^b will give you its sum.

Joginder Singh said...

Good thing boss

Unknown said...

void main()
{
int a=10,b=10,c;
clrscr();
c=a- -b;
printf("%d",c);
}

Manan Singla said...

You have used "+" operator (A++ which is A = A + 1) which is forbidden in the question itself.

Unknown said...

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Numerous global companies have set up offices in Amman- Jordan in a bid to promote partnerships and joint ventures with Middle East companies. NCITSolutions maintains a physical presence with staff on the ground in Amman- Jordan. We can also provide businesses with space and easy access to our business support services through our affiliate in Amman Jordan.

Unknown said...

Simply we can do it as:
#include

int main(){

int a,b;

printf("Enter any two integers: ");
scanf("%d%d",&a,&b);

for(int i=0;i<b;i++)
{
a++;
}

printf("Sum of two integers: %d",a);

return 0;
}

Unknown said...

+ is different and ++ is different... question says without using addition operator.. ++ is not addition operator. so the solution is correct. :)

Krishna said...

~b = -(b + 1 )
This is the right ones complement .

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