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Multiply any Number with using Bitwise Operator in C++
In this tutorial, we are going write a program that multiplies the given two numbers using bitwise operators.
The left shift (<<) operator is used for the multiplication whereas the right shift (>>) is used for the division.
The multiplication of two numbers x, y can be written as x * y = (x * 2) * (y / 2) if y is even else it's equal to x * y = (x * y) * (y / 2) + x.
So whenever the second number becomes odd, add the first number to the result. Let's see the steps to solve the problem.
Algorithm
- Initialise two numbers.
- Write a loop that iterates till the second number becomes 0.
- If the second number is odd, then add the first number to the result.
- Left shift first number by 1 bit.
- Right shift second number by 1 bit.
Implementation
Following is the implementation of the above algorithm in C++
#include <bits/stdc++.h>
using namespace std;
int multiplyTwoNumbers(int a, int b) {
int result = 0;
while (b > 0) {
if (b & 1) {
result += a;
}
a = a << 1;
b = b >> 1;
}
return result;
}
int main() {
cout << multiplyTwoNumbers(75, 4) << endl;
cout << multiplyTwoNumbers(90, 9) << endl;
cout << multiplyTwoNumbers(83, 66) << endl;
return 0;
}Output
If you run the above code, then you will get the following result.
300 810 5478
Updated on: 25-Oct-2021
8K+ Views
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