Can any one add two numbers without using + operator?

Answer 1

#include<stdio.h>

int main() {
int num1 = 16,
num2 = 14,
i;
//Result should be 30

while (num2 > 0) {
num1++;
num2--;
}
printf("%d", num1);
return (0);
}

In this example, I use the while loop to do it. " #include<stdio.h> " allows me to use "printf" which is the same as using 'cout' to print out text, but I prefer to use printf

By the way, this is for C++, not really for Java

Answer 2

#include<stdio.h>
//Insert main here

int y = 4;
int x = 6; //we are going to do y+x
int z = -1;
while (!(x==-6))
{
x--;
}
x = x-y;
int Result = x/z;
printf("%d", Result);
return (0);
//outputs 10. No plus operator used, only minus

Answer 3

nope + is used still... I don't wanna see + operator at all

Answer 4

int no1,no2;

while(no1--)
no2++;

cout<<no2;

//use it only if no1 is +ve
☺

Answer 5

Nope. if you use decrement(++),it also use + operator😂

Answer 6

Yes! it's possible via Binary Operators and bitwise operators.
Check below for code.

Answer 7

// Half Adder logic - Recursive approach
// Not my implementation

int add(int a,int b)
{
int sum,carry;
if (b == 0)
return a;
sum = a ^ b; // add without carrying
carry = (a & b) << 1; // carry, but don’t add
return add(sum,carry); // recurse
}

Answer 8

//let the two numbers are :
int n1,n2;

//then

while(n1--)
n2=-(-n2-1);

//👉n2 is the required sum
//without using +or++ operator
court<<n2;