更改uniform_int_distribution的范围

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)

Answer 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);
}

Answer 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);

Answer 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;
}