使用推力::reduce 计算 8 位整数向量的和而不会溢出

Question

I've got a device vector of type uint8_t and I want to compute a sum over it using thrust::reduce if possible. The problem is that I get overflow, since the sum will be much larger than 255. I thought the code below would compute the sum, by storing the results as 32 bit integers, but it doesn't seem to be the case. Is there a good way to accomplish this?

uint8_t * flags_d;
...
const int32_t N_CMP_BLOCKS = thrust::reduce( 
    thrust::device_pointer_cast( flags_d ), 
    thrust::device_pointer_cast( flags_d ) + N,
    (int32_t) 0,
    thrust::plus<int32_t>() );

Answer 1

I think the only solution that will work is to use thrust::transform_reduce to explicitly cast the 8 bit input data to a 32 bit quantity before the accumulation operation in the reduction. So I would expect something like this:

#include <thrust/transform_reduce.h>
#include <thrust/functional.h>
#include <thrust/execution_policy.h>

template<typename T1, typename T2>
struct char2int
{
  __host__ __device__ T2 operator()(const T1 &x) const
  {
    return static_cast<T2>(x);
  }
};

int main()
{
  unsigned char data[6] = {128, 100, 200, 102, 101, 123};
  int result = thrust::transform_reduce(thrust::host,
                                        data, data + 6,
                                        char2int<unsigned char,int>(),
                                        0,
                                        thrust::plus<int>());

  std::cout << "Result is " << result << std::endl;
 
  return 0;
}

to be more like what you had in mind.