Sum 2 Num Without Use Of Arithmetic Operators
Add two numbers without using arithmetic operators
Write a function Add() that returns sum of two integers. The function should not use any of the arithmetic operators (+, ++, –, -, .. etc).
Sum of two bits can be obtained by performing XOR (^) of the two bits. Carry bit can be obtained by performing AND (&) of two bits.
Above is simple Half Adder logic that can be used to add 2 single bits. We can extend this logic for integers. If x and y don’t have set bits at same position(s), then bitwise XOR (^) of x and y gives the sum of x and y. To incorporate common set bits also, bitwise AND (&) is used. Bitwise AND of x and y gives all carry bits. We calculate (x & y) << 1 and add it to x ^ y to get the required result.
Sum of two bits can be obtained by performing XOR (^) of the two bits. Carry bit can be obtained by performing AND (&) of two bits.
Above is simple Half Adder logic that can be used to add 2 single bits. We can extend this logic for integers. If x and y don’t have set bits at same position(s), then bitwise XOR (^) of x and y gives the sum of x and y. To incorporate common set bits also, bitwise AND (&) is used. Bitwise AND of x and y gives all carry bits. We calculate (x & y) << 1 and add it to x ^ y to get the required result.
#include<stdio.h>int Add(int x, int y){ // Iterate till there is no carry while (y != 0) { // carry now contains common set bits of x and y int carry = x & y; // Sum of bits of x and y where at least one of the bits is not set x = x ^ y; // Carry is shifted by one so that adding it to x gives the required sum y = carry << 1; } return x;}int main(){ printf("%d", Add(-10, 25)); return 0;} |
int Add(int x, int y){ if (y == 0) return x; else return Add( x ^ y, (x & y) << 1);} |
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