c++ - 如何在CUDA中将数据从unsigned int复制到ulong4

Question

.h文件:

#define VECTOR_SIZE 1024   

.cpp 文件:

int main ()
{
    unsigned int* A;
    A = new unsigned int [VECTOR_SIZE];

    CopyToDevice (A);
}

.cu 文件:

void CopyToDevice (unsigned int *A)
{
    ulong4 *UA
    unsigned int VectorSizeUlong4 = VECTOR_SIZE / 4;
    unsigned int VectorSizeBytesUlong4 = VectorSizeUlong4 * sizeof(ulong4);

    cudaMalloc( (void**)&UA, VectorSizeBytesUlong4 );

    // how to use cudaMemcpy to copy data from A to UA?

    // I tried to do the following but it gave access violation error:
    for (int i=0; i<VectorSizeUlong4; ++i)
    {
        UA[i].x = A[i*4 + 0];
        UA[i].y = A[i*4 + 1];
        UA[i].z = A[i*4 + 2];
        UA[i].w = A[i*4 + 3];
    }
    // I also tried to copy *A to device and then work on it instead going back to CPU to access *A every time but this did not work again
}

Answer 1

CUDA ulong4 是一个 16 字节对齐的结构，定义为

struct __builtin_align__(16) ulong4
{
  unsigned long int x, y, z, w;
};

这意味着您要用来填充 ulong4 流的四个连续 32 位无符号源整数流的大小相同。最简单的解决方案就包含在您发布的图像的文本中 - 只需将 unsigned int 指针转换(隐式或显式)到 ulong4 指针，使用 cudaMemcpy 直接在主机和设备内存上，并将生成的设备指针传递给您拥有的任何需要 ulong4 输入的内核函数。您的设备传输功能可能类似于:

ulong4* CopyToDevice (unsigned int* A)
{
    ulong4 *UA, *UA_h;
    size_t VectorSizeUlong4 = VECTOR_SIZE / 4;
    size_t VectorSizeBytesUlong4 = VectorSizeUlong4 * sizeof(ulong4);

    cudaMalloc( (void**)&UA, VectorSizeBytesUlong4);
    UA_h = reinterpret_cast<ulong4*>(A); // not necessary but increases transparency
    cudaMemcpy(UA, UA_h, VectorSizeBytesUlong4);

    return UA;   
}

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