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Home/ Questions/Q 1049815
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Editorial Team
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Editorial Team
Asked: May 16, 20262026-05-16T16:40:31+00:00 2026-05-16T16:40:31+00:00

I just started experimenting cuda with the following cude #include macro.hpp #include <algorithm> #include

  • 0

I just started experimenting cuda with the following cude

#include "macro.hpp"
#include <algorithm>
#include <iostream>
#include <cstdlib>

//#define double float
//#define double int

int RandomNumber(){return static_cast<double>(rand() % 1000);}

__global__ void sum3(double const* a,
             double const* b,
             double const* c,
             double * result, 
             unsigned const* n)
{    
   unsigned i = blockIdx.x;
   while(i < (*n))
{
  result[i] = (a[i] + b[i] + c[i]);
}
};


int main()
{

  static unsigned size = 1e2;
  srand(0);
  double* a = new double[size];
  double* b = new double[size];
  double* c = new double[size];
  double* result = new double[size];

  std::generate(a, a+size, RandomNumber);
  std::generate(b, b+size, RandomNumber);
  std::generate(c, c+size, RandomNumber);

  double* ad, *bd,* cd;
  double* resultd;

  unsigned * sized;
  std::cout << cudaMalloc((void**) &ad, size*sizeof(double)) << std::endl;
  std::cout << cudaMalloc((void**) &bd, size*sizeof(double)) << std::endl;
  std::cout << cudaMalloc((void**) &cd, size*sizeof(double)) << std::endl;
  std::cout << cudaMalloc((void**) &resultd, size*sizeof(double)) << std::endl;
  std::cout << cudaMalloc((void**) &sized, sizeof(unsigned)) << std::endl;

  cudaMemcpy((void**) &sized, &size, sizeof(unsigned), cudaMemcpyHostToDevice);

  //  print_array(a, size);
  for(int i = 0; i < 1000; ++i)
    {
      cudaMemcpy(ad, a, size*sizeof(double), cudaMemcpyHostToDevice);
      cudaMemcpy(bd, b, size*sizeof(double), cudaMemcpyHostToDevice);
      cudaMemcpy(cd, c, size*sizeof(double), cudaMemcpyHostToDevice);      
      sum3<<<size, 1>>>(ad, bd, cd, resultd, sized);
      cudaMemcpy(result, resultd, size*sizeof(double), cudaMemcpyDeviceToHost);
    }

#ifdef PRINT
  for( int i = 0; i < size; ++i)
    {
      std::cout << a[i] << ", "<< b[i] <<"," << c[i] << "," << result[i]<< std::endl;
    }
#endif

  cudaFree(ad);
  cudaFree(bd);
  cudaFree(cd);
  cudaFree(resultd);

  delete[] a;
  delete[] b;
  delete[] c;
  delete[] result;

  return 0;
}

Compile this on mac book pro without any problem. However when I try to run this I get

930, 22,538,899
691, 832,205,23
415, 655,148,120
872, 876,481,985
761, 909,583,619
841, 104,466,917
610, 635,911,52
//more useless numbers

I have compared my samples with the one in Cuda By Example and I dont see any major difference except type. Any pointer on this problem is appreciated.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-16T16:40:32+00:00Added an answer on May 16, 2026 at 4:40 pm
    while(i < (*n))
    {
      result[i] = (a[i] + b[i] + c[i]);
    }
    

    is wrong (infinite)

    this is wrong

    cudaMemcpy((void**) &sized, &size, sizeof(unsigned), cudaMemcpyHostToDevice);
    

    &sized is address of pointer variable, not pointer value

    Single number can be passed to device on the stack, so use

    unsigned size
    

    check return status of your cuda functions, http://www.drdobbs.com/high-performance-computing/207603131

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