cuda - 如何使用 PyCUDA 的 `prepare` 函数

Question

我在将正确的参数传递给 prepare function 时遇到问题(和prepared_call)到allocate of shared memory in PyCUDA .我以这种方式理解错误消息，我传递给 PyCUDA 的变量之一是 long而不是我想要的 float32 .但我看不到变量来自哪里。

此外，在我看来，official example和 documentation of prepare 如果 block 相互矛盾需要None或不。

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()

输出

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，我花了一段时间才找到答案，所以就这样吧。错误消息的原因是 data_gpu 是一个 GPUArray 实例，即你用

data_gpu = gpuarray.to_gpu(data)

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

此外，将块参数传递给 prepare 现在是 deprecated - 所以正确的调用应该是这样的:

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