Performing Arithmetic on Bitsets - C++ STL

C++ examples for STL:bitset

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Description

Performing Arithmetic on Bitsets

Demo Code

#include <stdexcept>
#include <bitset>

bool fullAdder(bool b1, bool b2, bool& carry) {
  bool sum = (b1 ^ b2) ^ carry;
  carry = (b1 && b2) || (b1 && carry) || (b2 && carry);
  return sum;/*  w w  w . java 2s .  c  o  m*/
}

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;
}

template<unsigned int N>
bool bitsetLtEq(const std::bitset<N>& x, const std::bitset<N>& y)
{
  for (int i=N-1; i >= 0; i--) {
    if (x[i] && !y[i]) return false;
    if (!x[i] && y[i]) return true;
  }
  return true;
}

template<unsigned int N>
bool bitsetLt(const std::bitset<N>& x, const std::bitset<N>& y)
{
  for (int i=N-1; i >= 0; i--) {
    if (x[i] && !y[i]) return false;
    if (!x[i] && y[i]) return true;
  }
  return false;
}

template<unsigned int N>
bool bitsetGtEq(const std::bitset<N>& x, const std::bitset<N>& y)
{
  for (int i=N-1; i >= 0; i--) {
    if (x[i] && !y[i]) return true;
    if (!x[i] && y[i]) return false;
  }
  return true;
}

template<unsigned int N>
bool bitsetGt(const std::bitset<N>& x, const std::bitset<N>& y)
{
  for (int i=N-1; i >= 0; i--) {
    if (x[i] && !y[i]) return true;
    if (!x[i] && y[i]) return false;
  }
  return false;
}

template<unsigned int N>
void bitsetAdd(std::bitset<N>& x, const std::bitset<N>& y)
{
  bool carry = false;
  for (int i = 0; i < N; i++) {
    x[i] = fullAdder(x[i], y[i], carry);
  }
}

template<unsigned int N>
void bitsetSubtract(std::bitset<N>& x, const std::bitset<N>& y) {
  bool borrow = false;
  for (int i = 0; i < N; i++) {
    if (borrow) {
      if (x[i]) {
        x[i] = y[i];
        borrow = y[i];
      }
      else {
        x[i] = !y[i];
        borrow = true;
      }
    }
    else {
      if (x[i]) {
        x[i] = !y[i];
        borrow = false;
      }
      else {
        x[i] = y[i];
        borrow = y[i];
      }
    }
  }
}

template<unsigned int N>
void bitsetMultiply(std::bitset<N>& x, const std::bitset<N>& y)
{
  std::bitset<N> tmp = x;
  x.reset();

  // we want to minimize the number of times we shift and add
  if (tmp.count() < y.count()) {
    for (int i=0; i < N; i++)
      if (tmp[i]) bitsetAdd(x, y << i);
  }
  else {
    for (int i=0; i < N; i++)
      if (y[i]) bitsetAdd(x, tmp << i);
  }
}

template<unsigned int N>
void bitsetDivide(std::bitset<N> x, std::bitset<N> y,std::bitset<N>& q, std::bitset<N>& r)
{
  if (y.none()) {
    throw std::domain_error("division by zero undefined");
  }
  q.reset();
  r.reset();
  if (x.none()) {
    return;
  }
  if (x == y) {
    q[0] = 1;
    return;
  }
  r = x;
  if (bitsetLt(x, y)) {
    return;
  }

  // count significant digits in divisor and dividend
  unsigned int sig_x;
  for (int i=N-1; i>=0; i--) {
    sig_x = i;
    if (x[i]) break;
  }
  unsigned int sig_y;
  for (int i=N-1; i>=0; i--) {
    sig_y = i;
    if (y[i]) break;
  }

  // align the divisor with the dividend
  unsigned int n = (sig_x - sig_y);
  y <<= n;

  // make sure the loop executes the right number of times
  n += 1;

  // long division algorithm, shift, and subtract
  while (n--)
  {
    // shift the quotient to the left
    if (bitsetLtEq(y, r))
    {
      // add a new digit to quotient
      q[n] = true;
      bitsetSubtract(r, y);
    }
    // shift the divisor to the right
    y >>= 1;
  }
}


#include <bitset>
#include <iostream>
#include <string>

using namespace std;

int main() {
  bitset<10> bits1(string("100010001"));
  bitset<10> bits2(string("000000011"));
  bitsetAdd(bits1, bits2);
  cout << bits1.to_string<char, char_traits<char>, allocator<char> >() << endl;
}

Result


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