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Arithmetic operations with std::bitset in C++

  • Last Updated : 17 Jun, 2021

A bitset is an array of boolean values, but each boolean value is not stored separately. Instead, bitset optimizes the space such that each bool takes 1-bit space only, so space taken by bitset say, bs is less than that of bool bs[N] and vector<bool> bs(N). However, a limitation of bitset is, N must be known at compile-time, i.e., a constant (this limitation is not there with vector and dynamic array)

Important Note:

  • Take care of integer overflow say if bitset is declared of size 3 and addition results 9, this is the case of integer overflow because 9 cannot be stored in 3 bits.
  • Take care for negative results as bitsets are converted to unsigned long integer, so negative numbers cannot be stored.

Addition of 2 bitsets: Follow the steps below to solve the problem:

  • Initialize a bool carry to false.
  • Create a bitset ans to store the sum of the two bitsets x and y.
  • Traverse the length of the bitsets x and y and use the fullAdder function to determine the value of the current bit in ans.
  • Return ans.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Utility function to add two bool values and calculate
// carry and sum
bool fullAdder(bool b1, bool b2, bool& carry)
{
    bool sum = (b1 ^ b2) ^ carry;
    carry = (b1 && b2) || (b1 && carry) || (b2 && carry);
    return sum;
}
// Function to add two bitsets
bitset<33> bitsetAdd(bitset<32>& x, bitset<32>& y)
{
    bool carry = false;
    // bitset to store the sum of the two bitsets
    bitset<33> ans;
    for (int i = 0; i < 33; i++) {
        ans[i] = fullAdder(x[i], y[i], carry);
    }
    return ans;
}
// Driver Code
int main()
{
    // Given Input
    bitset<32> a(25);
    bitset<32> b(15);
  
    // Store the result of addition
    bitset<33> result = bitsetAdd(a, b);
  
    cout << result;
    return 0;
}
Output



000000000000000000000000000101000

Time Complexity: O(N), N is length of bitset
Auxiliary Space: O(N)

Subtraction of 2 bitsets: Follow the steps below to solve the problem:

  • Initialize a bool borrow to false.
  • Create a bitset ans to store the difference between the two bitsets x and y.
  • Traverse the length of the bitsets x and y and use the fullSubtractor function to determine the value of the current bit in ans.
  • Return ans.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Utility function to subtract two bools and calculate diff
// and borrow
bool fullSubtractor(bool b1, bool b2, bool& borrow)
{
    bool diff;
    if (borrow) {
        diff = !(b1 ^ b2);
        borrow = !b1 || (b1 && b2);
    }
    else {
        diff = b1 ^ b2;
        borrow = !b1 && b2;
    }
    return diff;
}
// Function to calculate difference between two bitsets
bitset<33> bitsetSubtract(bitset<32> x, bitset<32> y)
{
    bool borrow = false;
    // bitset to store the sum of the two bitsets
    bitset<33> ans;
    for (int i = 0; i < 32; i++) {
        ans[i] = fullSubtractor(x[i], y[i], borrow);
    }
    return ans;
}
// Driver Code
int main()
{
    // Given Input
    bitset<32> a(25);
    bitset<32> b(15);
  
    // Store the result of addition
    bitset<33> result = bitsetSubtract(a, b);
  
    cout << result;
    return 0;
}
Output
000000000000000000000000000001010

Time Complexity: O(N), N is length of bitset
Auxiliary Space: O(N)

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