Generating random numbers with uniform distribution using Thrust

Question

I need to generate a vector with random numbers between 0.0 and 1.0 using Thrust. The only documented example I could find produces ve

Answer 1

Thrust has random generators you can use to produce sequences of random numbers. To use them with a device vector you will need to create a functor which returns a different element of the random generator sequence. The most straightforward way to do this is using a transformation of a counting iterator. A very simple complete example (in this case generating random single precision numbers between 1.0 and 2.0) could look like:

#include 
#include 
#include 
#include 
#include 

struct prg
{
    float a, b;

    __host__ __device__
    prg(float _a=0.f, float _b=1.f) : a(_a), b(_b) {};

    __host__ __device__
        float operator()(const unsigned int n) const
        {
            thrust::default_random_engine rng;
            thrust::uniform_real_distribution dist(a, b);
            rng.discard(n);

            return dist(rng);
        }
};


int main(void)
{
    const int N = 20;

    thrust::device_vector numbers(N);
    thrust::counting_iterator index_sequence_begin(0);

    thrust::transform(index_sequence_begin,
            index_sequence_begin + N,
            numbers.begin(),
            prg(1.f,2.f));

    for(int i = 0; i < N; i++)
    {
        std::cout << numbers[i] << std::endl;
    }

    return 0;
}

In this example, the functor prg takes the lower and upper bounds of the random number as an argument, with (0.f,1.f) as the default. Note that in order to have a different vector each time you call the transform operation, you should used a counting iterator initialised to a different starting value.