Vary range of uniform_int_distribution

Question

So i have a Random object:

typedef unsigned int uint32;

class Random {
public:
    Random() = default;
    Random(std::mt19937::result_type seed) : eng(seed) {}

private:
    uint32 DrawNumber();
    std::mt19937 eng{std::random_device{}()};
    std::uniform_int_distribution<uint32> uniform_dist{0, UINT32_MAX};
};

uint32 Random::DrawNumber()
{
    return uniform_dist(eng);
}

What's the best way I can vary (through another function or otherwise) the upper bound of of the distribution?

(also willing to take advice on other style issues)

Solution 1

Distribution objects are lightweight. Simply construct a new distribution when you need a random number. I use this approach in a game engine, and, after benchmarking, it's comparable to using good old rand().

Also, I've asked how to vary the range of distribution on GoingNative 2013 live stream, and Stephen T. Lavavej, a member of the standard committee, suggested to simply create new distributions, as it shouldn't be a performance issue.

Here's how I would write your code:

using uint32 = unsigned int;

class Random {
public:
    Random() = default;
    Random(std::mt19937::result_type seed) : eng(seed) {}
    uint32 DrawNumber(uint32 min, uint32 max);

private:        
    std::mt19937 eng{std::random_device{}()};
};

uint32 Random::DrawNumber(uint32 min, uint32 max)
{
    return std::uniform_int_distribution<uint32>{min, max}(eng);
}

Solution 2

You can simply create a std::uniform_int_distribution<uint32>::param_type and modify the range using the param() method. You can cut down the template noise with decltype :

decltype(uniform_dist.param()) new_range (0, upper);
uniform_dist.param(new_range);

Solution 3

I'm making the DrawNumber function public for my example. You can provide an overload that takes an upper bound, and then pass a new uniform_int_distribution::param_type to uniform_int_distribution::operator()

The param_type can be constructed using the same arguments as the corresponding distribution.

From N3337, §26.5.1.6/9 [rand.req.dist]

For each of the constructors of D taking arguments corresponding to parameters of the distribution, P shall have a corresponding constructor subject to the same requirements and taking arguments identical in number, type, and default values. Moreover, for each of the member functions of D that return values corresponding to parameters of the distribution, P shall have a corresponding member function with the identical name, type, and semantics.

where D is the type of a random number distribution function object and P is the type named by D's associated param_type

#include <iostream>
#include <random>

typedef unsigned int uint32;

class Random {
public:
    Random() = default;
    Random(std::mt19937::result_type seed) : eng(seed) {}

    uint32 DrawNumber();
    uint32 DrawNumber(uint32 ub);

private:
    std::mt19937 eng{std::random_device{}()};
    std::uniform_int_distribution<uint32> uniform_dist{0, UINT32_MAX};
};

uint32 Random::DrawNumber()
{
    return uniform_dist(eng);
}

uint32 Random::DrawNumber(uint32 ub)
{
    return uniform_dist(eng, decltype(uniform_dist)::param_type(0, ub));
}

int main()
{
  Random r;
  std::cout << r.DrawNumber() << std::endl;
  std::cout << r.DrawNumber(42) << std::endl;
}