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         


        

Answer 1

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 
#include 

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 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;
}