Question

我是CUDA和C的新手，我可以在以下方面提供一些帮助：我想将Cpu的GpuMats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

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)

我将非常感谢您的帮助！

Answer 1

我找到了解决问题的方法，方法是通过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);

CUDA将GpuMat的c数组传递给内核

1 个答案: