numbapro中的cuda代码错误

时间:2013-08-09 16:03:22

标签: python numpy cuda anaconda numba-pro

import numpy
import numpy as np
from numbapro import cuda


@cuda.autojit
def foo(aryA, aryB,out):
    d_ary1 = cuda.to_device(aryA)
    d_ary2 = cuda.to_device(aryB)
    #dd = numpy.empty(10, dtype=np.int32)
    d_ary1.copy_to_host(out)


griddim = 1, 2
blockdim = 3, 4
aryA = numpy.arange(10, dtype=np.int32)
aryB = numpy.arange(10, dtype=np.int32)
out = numpy.empty(10, dtype=np.int32)

foo[griddim, blockdim](aryA, aryB,out)
  

异常:由输入行11引起:   只能从全局变量,复数或数组中获取属性

我是numbapro的新手,需要提示!

1 个答案:

答案 0 :(得分:2)

@cuda.autotjit标记并编译foo()作为CUDA内核。内存传输操作应放在内核之外。它应该类似于以下代码:

import numpy
from numbapro import cuda

@cuda.autojit
def foo(aryA, aryB ,out):
    # do something here
    i = cuda.threadIdx.x + cuda.blockIdx.x * cuda.blockDim.x
    out[i] = aryA[i] + aryB[i]

griddim = 1, 2
blockdim = 3, 4
aryA = numpy.arange(10, dtype=numpy.int32)
aryB = numpy.arange(10, dtype=numpy.int32)
out = numpy.empty(10, dtype=numpy.int32)

# transfer memory
d_ary1 = cuda.to_device(aryA)
d_ary2 = cuda.to_device(aryB)
d_out = cuda.device_array_like(aryA) # like numpy.empty_like() but for GPU
# launch kernel
foo[griddim, blockdim](aryA, aryB, d_out)

# transfer memory device to host
d_out.copy_to_host(out)

print out

我建议新的NumbaPro用户查看https://github.com/ContinuumIO/numbapro-examples中的示例。