Using boost::random as the RNG for std::random_shuffle

问题

I have a program that uses the mt19937 random number generator from boost::random. I need to do a random_shuffle and want the random numbers generated for this to be from this shared state so that they can be deterministic with respect to the mersenne twister's previously generated numbers.

I tried something like this:

void foo(std::vector<unsigned> &vec, boost::mt19937 &state)
{
    struct bar {
        boost::mt19937 &_state;
        unsigned operator()(unsigned i) {
            boost::uniform_int<> rng(0, i - 1);
            return rng(_state);
        }
        bar(boost::mt19937 &state) : _state(state) {}
    } rand(state);

    std::random_shuffle(vec.begin(), vec.end(), rand);
}

But i get a template error calling random_shuffle with rand. However this works:

unsigned bar(unsigned i)
{
    boost::mt19937 no_state;
    boost::uniform_int<> rng(0, i - 1);
    return rng(no_state);
}
void foo(std::vector<unsigned> &vec, boost::mt19937 &state)
{
    std::random_shuffle(vec.begin(), vec.end(), bar);
}

Probably because it is an actual function call. But obviously this doesn't keep the state from the original mersenne twister. What gives? Is there any way to do what I'm trying to do without global variables?

回答1:

In C++03, you cannot instantiate a template based on a function-local type. If you move the rand class out of the function, it should work fine (disclaimer: not tested, there could be other sinister bugs).

This requirement has been relaxed in C++0x, but I don't know whether the change has been implemented in GCC's C++0x mode yet, and I would be highly surprised to find it present in any other compiler.

回答2:

In the comments, Robert Gould asked for a working version for posterity:

#include <algorithm>
#include <functional>
#include <vector>
#include <boost/random.hpp>

struct bar : std::unary_function<unsigned, unsigned> {
    boost::mt19937 &_state;
    unsigned operator()(unsigned i) {
        boost::uniform_int<> rng(0, i - 1);
        return rng(_state);
    }
    bar(boost::mt19937 &state) : _state(state) {}
};

void foo(std::vector<unsigned> &vec, boost::mt19937 &state)
{
    bar rand(state);
    std::random_shuffle(vec.begin(), vec.end(), rand);
}

回答3:

I'm using tr1 instead of boost::random here, but should not matter much.

The following is a bit tricky, but it works.

#include <algorithm>
#include <tr1/random>


std::tr1::mt19937 engine;
std::tr1::uniform_int<> unigen;
std::tr1::variate_generator<std::tr1::mt19937, 
                            std::tr1::uniform_int<> >gen(engine, unigen);
std::random_shuffle(vec.begin(), vec.end(), gen);

回答4:

I thought it was worth pointing out that this is now pretty straightforward in C++11 using only the standard library:

#include <random>
#include <algorithm>

std::random_device rd;
std::mt19937 randEng(rd());
std::shuffle(vec.begin(), vec.end(), randEng);

来源：https://stackoverflow.com/questions/147391/using-boostrandom-as-the-rng-for-stdrandom-shuffle

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c++

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boost-random