c++ - CUDA 将 GpuMat 的 c 数组传递给内核

Question

我是 CUDA 和 C 的新手，我可以在以下方面使用一些帮助: 我想将 GpuMats 的 C 数组传递给 CUDA 内核:

这是我的内核代码:

__global__
    void disparityFromDiffMapsKernel(cuda::PtrStepSzi* differenceMapsArray,
                                 int arraySize,
                                 cuda::PtrStepSzi disparityMap){
    int x = blockIdx.x * blockDim.x + threadIdx.x;
    int y = blockIdx.y * blockDim.y + threadIdx.y;



    //check if thread is inside the image
    if(x > differenceMapsArray[0].cols || y > differenceMapsArray[0].rows){
        return;
    }

    //do stuff

}

这是我初始化数组并调用内核的代码:

cuda::PtrStepSzi diffMaps[diffMapsSize];
for(int i = 0; i <= offset; i++){
    cuda::GpuMat diffMap(leftImageGPU.size(),CV_32SC1);
    cuda::PtrStepSzi diffMapPtr = diffMap;
    diffMaps[i] = diffMapPtr;
}

disparityFromDiffMapsKernel<<<numBlocks,threadsPerBlock>>>(diffMaps,diffMapsSize,disparityImageGPU); //gpu mat is initialized before

当我运行此代码时，出现以下 opencv 错误:

OpenCV(3.4.1) Error: Gpu API call (an illegal memory access was encountered)

我非常感谢任何帮助!

Answer 1

我找到了解决问题的方法，通过 cudaMalloc 和 cudaMemcpy 将数组移动到 GPU 内存(感谢 @sgarizvi 的提示)

这是最终代码，以防有人遇到类似问题:

// reserve memory for the diffmap ptrs arrays
cuda::PtrStepSzi* cpuDiffMapPtrs;
cpuDiffMapPtrs = (cuda::PtrStepSzi*) malloc(diffMapsSize * sizeof(cuda::PtrStepSzi));

cuda::PtrStepSzi* gpuDiffMapPtrs;
cudaMalloc(&gpuDiffMapPtrs, diffMapsSize * sizeof(cuda::PtrStepSzi));

//fill cpu array with ptrs to gpu mats
for(int i = 0; i< diffMapsSize; i++){
    cuda::GpuMat diffMap(leftImageGPU.size(),CV_32SC1);
    //do stuff with the gpu mats
    cpuDiffMapPtrs[i] = diffMap;
}

//copy cpu array to gpu
cudaMemcpy(gpuDiffMapPtrs,cpuDiffMapPtrs,diffMapsSize * sizeof(cuda::PtrStepSzi), cudaMemcpyHostToDevice);



disparityFromDiffMapsKernel<<<numBlocks,threadsPerBlock>>>(gpuDiffMapPtrs,diffMapsSize,halfKernelSize,disparityImageGPU);

// free the allocated memory
cudaFree(gpuDiffMapPtrs);
free(cpuDiffMapPtrs);