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Representing Large Fixed-Width Integers in C++

Large integers are common in many applications. In cryptography, for example, integers of 1,000 bits and larger are not uncommon. The implementation I presented here is based on a binary representation of numbers using a bitset, at a cost of some performance. What I lost in performance I more than made up for in simplicity. A more efficient implementation of arbitrary precision numbers could easily fill the book.

The BigInt template below uses the bitset from the header to (from my last post) allow you to represent unsigned integers using a fixed number of bits specified as a template parameter.


//big_int.hpp
#ifndef BIG_INT_HPP
#define BIG_INT_HPP
#include <bitset>
#include "bitset_arithmetic.hpp" // from my last post of bitset manipulation.
template<unsigned int N>
class BigInt

{
typedef BigInt self;
public:
BigInt( ) : bits( ) { }
BigInt(const self& x) : bits(x.bits) { }
BigInt(unsigned long x) {
int n = 0;
while (x) {
bits[n++] = x & 0x1;
x >>= 1;
}
}
explicit BigInt(const std::bitset& x) : bits(x) { }
// public functions
bool operator[](int n) const { return bits[n]; }
unsigned long toUlong( ) const { return bits.to_ulong( ); }
// operators
self& operator<<=(unsigned int n) {
bits <<= n;
return *this;
}
self& operator>>=(unsigned int n) {
bits >>= n;
return *this;
}
self operator++(int) {
self i = *this;
operator++( );
return i;
}
self operator--(int) {
self i = *this;
operator--( );
return i;
}
self& operator++( ) {
bool carry = false;
bits[0] = fullAdder(bits[0], 1, carry);
for (int i = 1; i < N; i++) {
bits[i] = fullAdder(bits[i], 0, carry);
}
return *this;
}
self& operator--( ) {
bool borrow = false;
bits[0] = fullSubtractor(bits[0], 1, borrow);
for (int i = 1; i < N; i++) {
bits[i] = fullSubtractor(bits[i], 0, borrow);
}
return *this;
}
self& operator+=(const self& x) {
bitsetAdd(bits, x.bits);
return *this;
}
self& operator-=(const self& x) {
bitsetSubtract(bits, x.bits);
return *this;
}
self& operator*=(const self& x) {
bitsetMultiply(bits, x.bits);
return *this;
}
self& operator/=(const self& x) {
std::bitset tmp;
bitsetDivide(bits, x.bits, bits, tmp);
return *this;
}
self& operator%=(const self& x) {
std::bitset tmp;
bitsetDivide(bits, x.bits, tmp, bits);
return *this;
}
self operator~( ) const { return ~bits; }
self& operator&=(self x) { bits &= x.bits; return *this; }
self& operator|=(self x) { bits |= x.bits; return *this; }
self& operator^=(self x) { bits ^= x.bits; return *this; }
// friend functions
friend self operator<<(self x, unsigned int n) { return x <<= n; }
friend self operator>>(self x, unsigned int n) { return x >>= n; }
friend self operator+(self x, const self& y) { return x += y; }
friend self operator-(self x, const self& y) { return x -= y; }
friend self operator*(self x, const self& y) { return x *= y; }
friend self operator/(self x, const self& y) { return x /= y; }
friend self operator%(self x, const self& y) { return x %= y; }
friend self operator^(self x, const self& y) { return x ^= y; }
friend self operator&(self x, const self& y) { return x &= y; }
friend self operator|(self x, const self& y) { return x |= y; }
// comparison operators
friend bool operator==(const self& x, const self& y) {
return x.bits == y.bits;
}
friend bool operator!=(const self& x, const self& y) {
return x.bits != y.bits;
}
friend bool operator>(const self& x, const self& y) {
return bitsetGt(x.bits, y.bits);
}
friend bool operator<(const self& x, const self& y) {
return bitsetLt(x.bits, y.bits);
}
friend bool operator>=(const self& x, const self& y) {
return bitsetGtEq(x.bits, y.bits);
}
friend bool operator<=(const self& x, const self& y) {
return bitsetLtEq(x.bits, y.bits);
}
private:
std::bitset<N> bits;
};

The BigInt template class could be used to represent factorials, as shown below,


//Using the big_int class
#include "big_int.hpp"
#include <iostream>
#include <vector>
#include <iterator>
#include <algorithm>
using namespace std;
void outputBigInt(BigInt<1024> x) {
vector<int> v;
if (x == 0) {
cout << 0;
return;
}
while (x > 0) {
v.push_back((x % 10).to_ulong( ));
x /= 10;
}
copy(v.rbegin( ), v.rend( ), ostream_iterator(cout, ""));
cout << endl;
}
int main( ) {
BigInt<1024> n(1);
// compute 32 factorial
for (int i=1; i <= 32; ++i) {
n *= i;
}
outputBigInt(n);
}

The program output is:
263130836933693530167218012160000000

 

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Performing Arithmetic on Bitsets in C++

The bitset class template comes with basic operations for manipulating sets of bits but doesn’t provide any arithmetic or comparison operations. This is because the library can’t safely assume what kind of numerical type a programmer might expect an arbitrary set of bits to represent.The functions below treat a bitset as a representation of an unsigned integer type, and provide you with functions for adding, subtracting, multiplying, dividing, and comparing them. These functions can provide a basis for writing custom-sized integer types. I did not use the most efficient algorithms I could for below function. Instead I chose the simplest possible algorithms because they are more easily understood. A much more efficient implementation would use similar algorithms, but would operate on words rather than single bits.

The functions below provide functions that allow arithmetic and comparison of bitset class template from the header as if it represents an unsigned integer.


#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;
}
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& x, const std::bitset& 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& x, const std::bitset& 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& x, const std::bitset& 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& x, const std::bitset& 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& x, const std::bitset& 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& x, const std::bitset& 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& x, const std::bitset& y)
{
std::bitset 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 x, std::bitset 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;
}
}

Using the bitset_arithmetic.hpp functions:

#include "bitset_arithmetic.hpp"
#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 >( ) << endl;
}

The program produces the following output:
0100010100