如何使用boost生成不同比特的随机多精度int

Question

我需要生成大量不同比特的随机多精度int(boost mpx_int(。我目前的方法是基于这两个例子：boost多精度随机，constexpr数组。为了以这种方式生成随机数，我需要将位数作为constexpr。我可以生成一个constexpr int数组，但后来我被卡住了，因为我无法从for循环中访问它们。

代码示例：

#include <boost/multiprecision/cpp_int.hpp>
#include <boost/random.hpp>
#include <iostream>
using namespace std;
using namespace boost::multiprecision;
using namespace boost::random;
template <int bit_limit>
struct N_bit_nums
{
constexpr N_bit_nums() : bits{}
{
for (int i = 0; i < bit_limit; ++i)
{
bits[i] = i + 1;
}
}
int bits[bit_limit];
};
int main()
{
constexpr int bit_limit = 3; // this will actually be on the order of 10^6
constexpr N_bit_nums<bit_limit> n_bit_nums{};
for (int i = 0; i < bit_limit; ++i)
{
independent_bits_engine<mt19937, n_bit_nums.bits[i], cpp_int> generator; // error: the value of ‘i’ is not usable in a constant expression
cpp_int rand_num = generator();
cout << rand_num << "n"; // just to see what is going on while testing
}
return 0;
}

Answer 1

我能够通过将independent_bits_engine固定为所需的最大位数，然后屏蔽为所需位数来实现这一点。

示例：

#include <boost/multiprecision/cpp_int.hpp>
#include <boost/random.hpp>
#include <iostream>
using namespace std;
using namespace boost::multiprecision;
using namespace boost::random;
int main()
{
constexpr int bit_limit = 100;
independent_bits_engine<mt19937, bit_limit, cpp_int> generator;
// prints random numbers of bit sizes from 1 to bit_limit
for (int n = 1; n <= bit_limit; n++)
{
cpp_int rand_num = generator();              // next random value
cpp_int n_bit_mask = pow(cpp_int{2}, n) - 1; // n bits mask
cpp_int n_bit_num = rand_num & n_bit_mask;   // take n lsb
cout << n_bit_num << "n";                   // print the n bit random number
}
return 0;
}