如何使用PyCUDA中的`prepare`函数

Question

我有问题将正确的参数传递给prepare函数(和到prepared_call)以在PyCUDA中分配共享内存.我以这种方式理解错误消息,我传递给PyCUDA的一个变量是一个long而不是我想要的float32.但我看不出变量来自哪里.

此外,在我看来,官方的例子和文件prepare相互矛盾,是否block需要None.

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

kernel_code ="""
__device__ void loadVector(float *target, float* source, int dimensions )
{
    for( int i = 0; i < dimensions; i++ ) target[i] = source[i];
}
__global__ void kernel(float* data, int dimensions, float* debug)
{
    extern __shared__ float mean[];
    if(threadIdx.x == 0) loadVector( mean, &data[0], dimensions );
    debug[threadIdx.x]=  mean[threadIdx.x];
}
"""

dimensions = 12
np.random.seed(23)
data = np.random.randn(dimensions).astype(np.float32)
data_gpu = gpuarray.to_gpu(data)
debug = gpuarray.zeros(dimensions, dtype=np.float32)

mod = compiler.SourceModule(kernel_code)
kernel = mod.get_function("kernel")
kernel.prepare("PiP",block = (dimensions, 1, 1),shared=data.size)
grid = (1,1)
kernel.prepared_call(grid,data_gpu,dimensions,debug)
print debug.get()

产量

Traceback (most recent call last):
File "shared_memory_minimal_example.py", line 28, in <module>
kernel.prepared_call(grid,data_gpu,dimensions,debug)
File "/usr/local/lib/python2.6/dist-packages/pycuda-0.94.2-py2.6-linux-x86_64.egg/pycuda/driver.py", line 230, in function_prepared_call
func.param_setv(0, pack(func.arg_format, *args))
pycuda._pvt_struct.error: cannot convert argument to long

Answer 1

我遇到了同样的问题,我花了一段时间才得出答案,所以这里就是这样.错误消息的原因是data_gpu是GPUArray实例,即您使用它

data_gpu = gpuarray.to_gpu(data)

要将其传递给prepared_call,您需要执行data_gpu.gpudata以获取关联的DeviceAllocation实例(即有效地指向设备内存位置的指针).

此外,现在不推荐将块参数传递给prepare - 所以正确的调用将是这样的:

data_gpu = gpuarray.to_gpu(data)
func.prepare( "P" )
grid = (1,1)
block = (1,1,1)
func.prepared_call( grid, block, data_gpu.gpudata )