为CUDA实现32位内存集的“正确”方法是什么？

Question

CUDA has the API call

cudaError_t cudaMemset (void *devPtr, int value, size_t count)

which fills a buffer with a single-byte value. I want to fill it with a multi-byte value. Suppose, for the sake of simplicity, that I want to fill devPtr with a 32-bit (4-byte) value, and suppose we can ignore endianness. Now, the CUDA driver has the following API call:

CUresult cuMemsetD32(CUdeviceptr dstDevice, unsigned int ui, size_t N)

So is it enough for me to just: obtain the CUdeviceptr from the device-memory-space pointer, then make the driver API call? Or is there something else I need to be doing?

Answer 1

As of about CUDA 3.0, runtime API device pointers (and everything else) are interoperable with the driver API. So yes, you can use cuMemsetD32 to fill a runtime API allocation with a 32 bit value. The size of CUdeviceptr will match the size of void * on you platform and it is safe to cast a pointer from the CUDA API to CUdeviceptr or vice versa .

Answer 2

Based on talonmies' answer , it seems a reasonable (though ugly) approach would be:

#include <stdint.h>
inline cudaError_t cudaMemsetTyped<T>(void *devPtr, T value, size_t count);

#define INSTANTIATE_CUDA_MEMSET_TYPED(_nbits) \
inline cudaError_t cudaMemsetTyped<int ## _nbits ## _t>(void *devPtr, int ## _nbits ## _t value, size_t count) { \
    cuMemsetD ## _nbits( reinterpret_cast<CUdeviceptr>(devPtr), value, count); \
} \
inline cudaError_t cudaMemsetTyped<uint ## _nbits ## _t>(void *devPtr, uint ## _nbits ## _t value, size_t count) { \
    cuMemsetD ## _nbits( reinterpret_cast<CUdeviceptr>(devPtr), reinterpret_cast<uint ## _nbits ## _t>(value), count); \
} \

INSTANTIATE_CUDA_MEMSET_TYPED(8)
INSTANTIATE_CUDA_MEMSET_TYPED(16)
INSTANTIATE_CUD_AMEMSET_TYPED(32)

#undef INSTANTIATE_CUDA_MEMSET_TYPED(_nbits)

inline cudaError_t cudaMemsetTyped<float>(void *devPtr, float value, size_t count) {
    cuMemsetD32( reinterpret_cast<CUdeviceptr>(devPtr), reinterpret_cast<int>(value), count);
}

(no cuMemset64 it seems, so no double either)