我用Cuda编写的全局函数仅运行最后一个块

Question

已解决：抱歉，这是我的错，我应该在内核函数中使用atomicAdd(times,1);而不是*times++。

我这样称呼内核函数

dim3 Dg(blockSize, blockSize, blockSize);
dim3 Db(8, 8, 8);
voxelize << < Dg, Db >> > ();
cudaDeviceSynchronize();

但是我发现我的程序只能解决部分问题，因此我在全局函数printf()中使用voxelize ()就像下面的代码

__global__ void voxelize(){
    printf("the thread blockIdx.x %d, blockIdx.y %d blockIdx.z %d\n", blockIdx.x, blockIdx.y, blockIdx.z);
    unsigned int xIndex = blockDim.x * blockIdx.x + threadIdx.x;
    unsigned int yIndex = blockDim.y * blockIdx.y + threadIdx.y;
    unsigned int zIndex = blockDim.z * blockIdx.z + threadIdx.z;
    unsigned int i = zIndex * blockDim.x*blockDim.y+ yIndex * blockDim.x+ xIndex;
}    

输出仅显示每个维的最后一部分（即blockIdx.x始终为5，只有一部分blockIndex.z从0更改为5）。但我不明白为什么，是当我调用此内核函数时有什么问题吗？ 我的计算机装有GTX1050Ti MaxQ和cuda 10。

---

之后，我向内核传递了一个指针以监视运行时间。

 int blockSize = ceil(pow(triangles.size() 69664 / 512.0, 1.0 / 3));
 dim3 Dg(blockSize, blockSize, blockSize);
 dim3 Db(8, 8, 8);
 int* times = new int(0);
 int* gpu_times;
 cudaMalloc((void **)&gpu_times, sizeof(int));
 cudaMemcpy(gpu_times, times, sizeof(int), cudaMemcpyHostToDevice);
 voxelize << < Dg, Db >> > (gpu_times);
 cudaDeviceSynchronize();
 cudaMemcpy(times, gpu_times, sizeof(int), cudaMemcpyDeviceToHost);
 std::cout << *times << std::endl;

内核被修改为

__global__ void voxelize(int* times){
    (*times)++;
    printf("the thread blockIdx.x %d, blockIdx.y %d blockIdx.z %d\n", blockIdx.x, blockIdx.y, blockIdx.z);
    unsigned int xIndex = blockDim.x * blockIdx.x + threadIdx.x;
    unsigned int yIndex = blockDim.y * blockIdx.y + threadIdx.y;
    unsigned int zIndex = blockDim.z * blockIdx.z + threadIdx.z;
    unsigned int i = zIndex * blockDim.x*blockDim.y+ yIndex * blockDim.x+ xIndex;
}    

输出为 输出显示它运行了141次，但实际上输出应该远远大于69664

---

对不起，这是我的错，我应该使用atomicAdd(times,1);而不是*times++。

但是为什么printf()只输出我之前所述的索引的一部分？

Answer 1

关于您的printf问题

您需要调用cudaDeviceSynchronize()（为清楚起见，省略了错误检查），如果您使用大量的printf（这种情况），则还需要cudaDeviceSetLimit(...)：

#include <stdio.h>

__global__ void voxelize(){
    printf("the thread blockIdx.x %d, blockIdx.y %d blockIdx.z %d\n", blockIdx.x, blockIdx.y, blockIdx.z);
    unsigned int xIndex = blockDim.x * blockIdx.x + threadIdx.x;
    unsigned int yIndex = blockDim.y * blockIdx.y + threadIdx.y;
    unsigned int zIndex = blockDim.z * blockIdx.z + threadIdx.z;
    unsigned int i = zIndex * blockDim.x*blockDim.y+ yIndex * blockDim.x+ xIndex;
}

int main()
{
  // Increase device printf buffer to 50 MiB
  cudaDeviceSetLimit(cudaLimitPrintfFifoSize, 50*1024*1024);
  dim3 Dg(5, 5, 5);
  dim3 Db(8, 8, 8);
  voxelize<<<Dg, Db>>>();
  cudaDeviceSynchronize();

  return 0;
}

这将打印如下内容：

the thread blockIdx.x 2, blockIdx.y 3 blockIdx.z 4
the thread blockIdx.x 2, blockIdx.y 3 blockIdx.z 4
the thread blockIdx.x 2, blockIdx.y 3 blockIdx.z 4
the thread blockIdx.x 2, blockIdx.y 3 blockIdx.z 4
the thread blockIdx.x 2, blockIdx.y 3 blockIdx.z 4
[...]

然后您可以像这样检查它：

# This will keep one line per block and count them, so 5*5*5 == 125
$ ./a.out | sort | uniq | wc -l
125

# This will output one line per thread and count them, so 5*5*5 * 8*8*8 == 64000
$ ./a.out | wc -l
64000

为您计算问题

您不能这样做：(*times)++;。您将遇到并发问题。您需要使用atomic functions。