3

我刚开始用以下 cude 试验 cuda

#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;
}

在 mac book pro 上编译这个没有任何问题。但是，当我尝试运行它时，我得到了

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

我已经将我的样本与 Cuda By Example 中的样本进行了比较，除了类型之外，我没有发现任何重大差异。对此问题的任何指针表示赞赏。

4

2 回答 2

1 
while(i < (*n))
{
  result[i] = (a[i] + b[i] + c[i]);
}

是错误的（无限）

这是错误的

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

&sized是指针变量的地址，而不是指针值

单个数字可以传递给堆栈上的设备，所以使用

unsigned size

检查您的 cuda 函数的返回状态，http://www.drdobbs.com/high-performance-computing/207603131

于 2010-09-05T01:11:29.667 回答
-1

你写了：

double* a = new double[size];

所以，“a”是一个指向双精度数组的指针，那么你说

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

这是错误的，你应该说

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

如果你说明你想让你的程序做什么，会更容易帮助你。

另外，你把

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

但你可以

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

取决于你想要做什么。

于 2010-09-05T02:53:32.263 回答