在不使用iostream的情况下保存c ++ 11随机生成器的状态

Question

在不使用iostream接口的情况下,存储C++ 11随机生成器状态的最佳方法是什么.我想像这里列出的第一个替代品那样做[1]？但是,此方法要求对象包含PRNG状态且仅包含PRNG状态.在partucular,它如果实现使用PIMPL模式失败(至少这是有可能重新加载状态,而不是用坏的数据加载它,当应用程序崩溃),或有与没有的PRNG对象相关联的多个状态变量与生成的序列有关.

对象的大小是实现定义的:

• g++ (tdm64-1) 4.7.1sizeof(std::mt19937)==2504但是给了
• Ideone http://ideone.com/41vY5j给出2500

我缺少像这样的成员函数

1. size_t state_size();
2. const size_t* get_state() const;
3. void set_state(size_t n_elems,const size_t* state_new);

(1)应返回随机生成器状态数组的大小

(2)应返回指向状态数组的指针.指针由PRNG管理.

(3)应从std::min(n_elems,state_size())state_new指向的缓冲区中复制缓冲区

这种接口允许更灵活的状态操作.或者是否有任何PRNG:s的状态不能表示为无符号整数数组？

[1] 比使用流来节省增强随机生成器状态更快

Answer 1

我已经为我在OP评论中提到的方法编写了一个简单的（-ish）测试。它显然没有经过实战考验，但这个想法已经得到体现——你应该能够从这里得到它。

由于读取的字节量比序列化整个引擎要少得多，因此两种方法的性能实际上可能相当。测试这个假设以及进一步的优化，留给读者作为练习。

#include <iostream>
#include <random>
#include <chrono>
#include <cstdint>
#include <fstream>

using namespace std;

struct rng_wrap
{
    // it would also be advisable to somehow
    // store what kind of RNG this is,
    // so we don't deserialize an mt19937
    // as a linear congruential or something,
    // but this example only covers mt19937

    uint64_t seed;
    uint64_t invoke_count;
    mt19937 rng;

    typedef mt19937::result_type result_type;

    rng_wrap(uint64_t _seed) :
        seed(_seed),
        invoke_count(0),
        rng(_seed)
    {}

    rng_wrap(istream& in) {
        in.read(reinterpret_cast<char*>(&seed), sizeof(seed));
        in.read(reinterpret_cast<char*>(&invoke_count), sizeof(invoke_count));
        rng = mt19937(seed);
        rng.discard(invoke_count);
    }

    void discard(unsigned long long z) {
        rng.discard(z);
        invoke_count += z;
    }

    result_type operator()() {
        ++invoke_count;
        return rng();
    }

    static constexpr result_type min() {
        return mt19937::min();
    }

    static constexpr result_type max() {
        return mt19937::max();
    }
};

ostream& operator<<(ostream& out, rng_wrap& wrap)
{
    out.write(reinterpret_cast<char*>(&(wrap.seed)), sizeof(wrap.seed));
    out.write(reinterpret_cast<char*>(&(wrap.invoke_count)), sizeof(wrap.invoke_count));
    return out;
}

istream& operator>>(istream& in, rng_wrap& wrap)
{
    wrap = rng_wrap(in);
    return in;
}

void test(rng_wrap& rngw, int count, bool quiet=false)
{
    uniform_int_distribution<int> integers(0, 9);
    uniform_real_distribution<double> doubles(0, 1);
    normal_distribution<double> stdnorm(0, 1);

    if (quiet) {
        for (int i = 0; i < count; ++i)
            integers(rngw);

        for (int i = 0; i < count; ++i)
            doubles(rngw);

        for (int i = 0; i < count; ++i)
            stdnorm(rngw);
    } else {
        cout << "Integers:\n";
        for (int i = 0; i < count; ++i)
            cout << integers(rngw) << " ";

        cout << "\n\nDoubles:\n";
        for (int i = 0; i < count; ++i)
            cout << doubles(rngw) << " ";

        cout << "\n\nNormal variates:\n";
        for (int i = 0; i < count; ++i)
            cout << stdnorm(rngw) << " ";
        cout << "\n\n\n";
    }
}


int main(int argc, char** argv)
{
    rng_wrap rngw(123456790ull);

    test(rngw, 10, true);  // this is just so we don't start with a "fresh" rng
    uint64_t seed1 = rngw.seed;
    uint64_t invoke_count1 = rngw.invoke_count;

    ofstream outfile("rng", ios::binary);
    outfile << rngw;
    outfile.close();

    cout << "Test 1:\n";
    test(rngw, 10);  // test 1

    ifstream infile("rng", ios::binary);
    infile >> rngw;
    infile.close();

    cout << "Test 2:\n";
    test(rngw, 10);  // test 2 - should be identical to 1

    return 0;
}