Question

我想使用nppiFilterGauss_8u_C1R，但是当我使用cuda-memcheck时，它总是报告越界，这是我的源代码：

Npp8u* p1 = NULL;
Npp8u* p2 = NULL;
unsigned char* p3 = NULL;
unsigned char* p4 = NULL;
int step1 = 0;
int step2 = 0;
NppiSize roi;
roi.width = 352*288;
roi.height = 1;
int ret = 0;

p1 = nppiMalloc_8u_C1(352, 288, &step1);
p2 = nppiMalloc_8u_C1(352, 288, &step2);
cudaMalloc((void**)&p3, 352*288);
cudaMalloc((void**)&p4, 352*288);

printf("p1[%x],p2[%x],p3[%x],p4[%x]\n", p1, p2, p3, p4);
printf("step1[%d]\n", step1);
printf("step2[%d]\n", step2);

int count = 1;
while(count < 3) {
  // ret = nppiFilterGauss_8u_C1R(p1, step1, p2, step2, roi, NPP_MASK_SIZE_3_X_3);
  ret = nppiFilterGauss_8u_C1R(p3, 352*288, p4, 352*288, roi, NPP_MASK_SIZE_3_X_3);
  printf("count[%d],ret[%d]\n", count, ret);
  if(ret) {
    break;
  }
  count++;
}

nppiFree(p1);
nppiFree(p2);
cudaFree(p3);
cudaFree(p4);

这是错误：

GPU Device 0: "GK20A" with compute capability 3.2
p1[ab84a000],p2[ab86e000],p3[ab892000],p4[ab8aac00]
step1[512]
step2[512]
count[1],ret[0]
count[2],ret[0]
========= CUDA-MEMCHECK
========= Invalid __global__ read of size 1
=========     at 0x00000448 in void ForEachTupleByteQuad<unsigned char, int=1, TupleByteQuadFunctor<unsigned char, int=1, FilterGauss3x3QuadNew<unsigned char, int=1>>>(Tuple8<unsigned char, int=1>*, int, NppiSize, unsigned char)
=========     by thread (31,0,0) in block (395,0,0)
=========     Address 0xab8c3800 is out of bounds
=========
========= Program hit cudaErrorLaunchFailure (error 4) due to "unspecified launch failure" on CUDA API call to cudaFree.
=========
========= Program hit cudaErrorLaunchFailure (error 4) due to "unspecified launch failure" on CUDA API call to cudaFree.
=========
========= Program hit cudaErrorLaunchFailure (error 4) due to "unspecified launch failure" on CUDA API call to cudaFree.
=========
========= Program hit cudaErrorLaunchFailure (error 4) due to "unspecified launch failure" on CUDA API call to cudaFree.
=========
========= ERROR SUMMARY: 5 errors*

有人可以解释正确的方法吗？

Answer 1

应用掩模大小为3x3的高斯滤波器需要从当前像素读取顶部/底部和左/右像素。这意味着在计算像素（0,0）的值时，实际上是从像素（-1，-1）读取的。为避免这种情况，您需要调整投资回报率或使用自动处理边界的NPP功能。

从您的代码中可以看出：

Npp8u* p1 = NULL;
Npp8u* p2 = NULL;

int step1 = 0;
int step2 = 0;
NppiSize roi;
roi.width = 352 - 2; //roi is two pixels smaller: one pixel removed left, one on right side
roi.height = 288 - 2; //same for height
int ret = 0;

p1 = nppiMalloc_8u_C1(352, 288, &step1); //use nppiMalloc and not cudaMalloc for best performance
p2 = nppiMalloc_8u_C1(352, 288, &step2); //(NPP uses internal heuristics which step size suits best...)

printf("p1[%x],p2[%x]\n", p1, p2);
printf("step1[%d]\n", step1);
printf("step2[%d]\n", step2);

int count = 1;
while (count < 3) {
    //move pointers from pixel (0,0) to pixel (1,1) = add one line step plus one, roi is two pixels smaller:
    ret = nppiFilterGauss_8u_C1R(p1 + step1 + 1, step1, p2 + step2 + 1, step2, roi, NPP_MASK_SIZE_3_X_3);
    printf("count[%d],ret[%d]\n", count, ret);
    if (ret) {
        break;
    }
    count++;
}

//Or use NPP function including border handling:
NppiPoint srcPoint;
srcPoint.x = 0;
srcPoint.y = 0;

roi.width = 352;
roi.height = 288;
ret = nppiFilterGaussBorder_8u_C1R(p1, step1, roi, srcPoint, p2, step2, roi, NPP_MASK_SIZE_3_X_3, NPP_BORDER_REPLICATE);

nppiFree(p1);
nppiFree(p2);

此代码通过CudaMemCheck没有任何麻烦。

cuda-memcheck报告nppiFilterGauss_8u_C1R超出范围

1 个答案: