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1661

answers:

3

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?

+7  A: 

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.

coppro
Tested moving the struct out, and works.
Chris Jester-Young
While you're moving the struct out to global level, feel free to also make it inherit from std::unary_function too. :-)
Chris Jester-Young
yes this works, thank you very much. I didn't think that moving the class definition out of the function would make a difference.
Greg Rogers
+6  A: 

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);
}
Chris Jester-Young
Chris Jester-Young
Out of curiosity what benefit does inheriting from unary_function give? It's obvious from operator() what the input and output are...
Greg Rogers
It makes the functors more composable. i.e., it provides some typedefs that make building other functors out of your functor easier. Let me hunt down a link for you....
Chris Jester-Young
I can't find an online article, however, in Effective STL, Item 40 ("Make functor classes adaptable") talks about this.
Chris Jester-Young
Thanks Chris Jester-Young!
Robert Gould
Couldnt you have just used hist original bar and then passed std::ptr_fun(bar) to the std::random_shuffle, seems like a lot less typing to do the same thing :-P
Evan Teran
A: 

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

The following is a bit tricky, but it works.

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);
baol