循环展开时出现“资源不足”错误

Question

当我将内核中的展开循环从 8 个增加到 9 个时，它会因 资源不足 错误而中断。

我读到 如何我是否诊断由于资源不足而导致 CUDA 启动失败？ 参数不匹配和寄存器过度使用可能是一个问题，但这里似乎不是这种情况。

我的内核计算 n 个点和 m 个质心之间的距离，并为每个点选择最近的质心。它适用于 8 维，但不适用于 9 维。当我设置 dimensions=9 并取消注释距离计算的两行时，我得到一个 pycuda._driver.LaunchError: cuLaunchGrid failed: launch out资源。

您认为什么可能导致这种行为？还有哪些其他原因会导致资源不足*？

我使用的是 Quadro FX580。这是最小的例子。为了在实际代码中展开，我使用模板。

import numpy as np
from pycuda import driver, compiler, gpuarray, tools
import pycuda.autoinit


## preference
np.random.seed(20)
points = 512
dimensions = 8
nclusters = 1

## init data
data = np.random.randn(points,dimensions).astype(np.float32)
clusters = data[:nclusters]

## init cuda
kernel_code = """

      // the kernel definition 
    __device__ __constant__ float centroids[16384];

    __global__ void kmeans_kernel(float *idata,float *g_centroids,
    int * cluster, float *min_dist, int numClusters, int numDim) {
    int valindex = blockIdx.x * blockDim.x + threadIdx.x ;
    float increased_distance,distance, minDistance;
    minDistance = 10000000 ;
    int nearestCentroid = 0;
    for(int k=0;k<numClusters;k++){
      distance = 0.0;
      increased_distance = idata[valindex*numDim] -centroids[k*numDim];
      distance = distance +(increased_distance * increased_distance);
      increased_distance =  idata[valindex*numDim+1] -centroids[k*numDim+1];
      distance = distance +(increased_distance * increased_distance);
      increased_distance =  idata[valindex*numDim+2] -centroids[k*numDim+2];
      distance = distance +(increased_distance * increased_distance);
      increased_distance =  idata[valindex*numDim+3] -centroids[k*numDim+3];
      distance = distance +(increased_distance * increased_distance);
      increased_distance =  idata[valindex*numDim+4] -centroids[k*numDim+4];
      distance = distance +(increased_distance * increased_distance);
      increased_distance =  idata[valindex*numDim+5] -centroids[k*numDim+5];
      distance = distance +(increased_distance * increased_distance);
      increased_distance =  idata[valindex*numDim+6] -centroids[k*numDim+6];
      distance = distance +(increased_distance * increased_distance);
      increased_distance =  idata[valindex*numDim+7] -centroids[k*numDim+7];
      distance = distance +(increased_distance * increased_distance);
      //increased_distance =  idata[valindex*numDim+8] -centroids[k*numDim+8];
      //distance = distance +(increased_distance * increased_distance);

      if(distance <minDistance) {
        minDistance = distance ;
        nearestCentroid = k;
        } 
      }
      cluster[valindex]=nearestCentroid;
      min_dist[valindex]=sqrt(minDistance);
    } 
 """
mod = compiler.SourceModule(kernel_code)
centroids_adrs = mod.get_global('centroids')[0]    
kmeans_kernel = mod.get_function("kmeans_kernel")
clusters_gpu = gpuarray.to_gpu(clusters)
cluster = gpuarray.zeros(points, dtype=np.int32)
min_dist = gpuarray.zeros(points, dtype=np.float32)

driver.memcpy_htod(centroids_adrs,clusters)

distortion = gpuarray.zeros(points, dtype=np.float32)
block_size= 512

## start kernel
kmeans_kernel(
    driver.In(data),driver.In(clusters),cluster,min_dist,
    np.int32(nclusters),np.int32(dimensions),
    grid = (points/block_size,1),
    block = (block_size, 1, 1),
)
print cluster
print min_dist

Answer 1

由于您的 block_size (512) 太大，您的寄存器已用完。

ptxas 通过注释行报告您的内核使用 16 个寄存器：

$ nvcc test.cu -Xptxas --verbose
ptxas info    : Compiling entry function '_Z13kmeans_kernelPfS_PiS_ii' for 'sm_10'
ptxas info    : Used 16 registers, 24+16 bytes smem, 65536 bytes cmem[0]

取消注释这些行会将寄存器使用增加到 17 个，并在运行时出现错误：

$ nvcc test.cu -run -Xptxas --verbose
ptxas info    : Compiling entry function '_Z13kmeans_kernelPfS_PiS_ii' for 'sm_10'
ptxas info    : Used 17 registers, 24+16 bytes smem, 65536 bytes cmem[0]
error: too many resources requested for launch

内核每个线程使用的物理寄存器数量限制了您可以在运行时启动的块。 SM 1.0 设备有 8K 个寄存器可供线程块使用。我们可以将其与内核的寄存器需求进行比较：17 * 512 = 8704 > 8K。在 16 个寄存器处，您原来的注释内核只是吱吱作响：16 * 512 = 8192 == 8K。

当未指定架构时，nvcc 默认为 SM 1.0 设备编译内核。 PyCUDA 可能以同样的方式工作。

要解决您的问题，您可以减小 block_size （例如，256）或找到一种方法来配置 PyCUDA 来为 SM 2.0 设备编译内核。 SM 2.0 设备（例如 QuadroFX 580）提供 32K 寄存器，对于您的原始 block_size 512 来说绰绰有余。