cuda - cublasSetVector() 与 cudaMemcpy()

Question

我想知道是否有区别:

// cumalloc.c - Create a device on the device
HOST float * cudamath_vector(const float * h_vector, const int m)
{
  float *d_vector = NULL;
  cudaError_t cudaStatus;
  cublasStatus_t cublasStatus;

  cudaStatus = cudaMalloc(&d_vector, sizeof(float) * m );

  if(cudaStatus == cudaErrorMemoryAllocation) {
    printf("ERROR: cumalloc.cu, cudamath_vector() : cudaErrorMemoryAllocation");
    return NULL;
  }


  /*    THIS: */ cublasSetVector(m, sizeof(*d_vector), h_vector, 1, d_vector, 1);

  /* OR THAT: */ cudaMemcpy(d_vector, h_vector, sizeof(float) * m, cudaMemcpyHostToDevice);


  return d_vector;
}

cublasSetVector()有两个参数 incx和 incy和 documentation says :

The storage spacing between consecutive elements is given by incx for the source vector x and for the destination vector y.

在 NVIDIA forum有人说:

iona_me: "incx and incy are strides measured in floats."

那么这是否意味着对于 incx = incy = 1 float[] 的所有元素将是 sizeof(float) -对齐并为 incx = incy = 2会有一个 sizeof(float) - 每个元素之间的填充？

除了这两个参数和 cublasHandle - 做cublasSetVector()还有什么cudaMalloc()不行吗？

传递 的向量/矩阵是否会保存？不是 用他们各自的 cublas*() 创建函数到其他 CUBLAS 函数来操作它们？

Answer 1

Massimiliano Fatica 提供的 thread of the NVIDIA Forum 中有一条评论，证实了我在上述评论中的陈述(或者，更好地说，我的评论源于对阅读我链接到的帖子的记忆)。尤其

cublasSetVector, cubblasGetVector, cublasSetMatrix, cublasGetMatrix are thin wrappers around cudaMemcpy and cudaMemcpy2D. Therefore, no significant performance differences are expected between the two sets of copy functions.

因此，您可以安全地将 cudaMalloc 创建的任何数组作为输入传递给 cublasSetVector 。

关于步伐，指南中可能有打印错误(从 CUDA 6.0 开始)，它说

The storage spacing between consecutive elements is given by incx for the source vector x and for the destination vector y.

但 也许 应该读作

The storage spacing between consecutive elements is given by incx for the source vector x and incy for the destination vector y.