如何使用jupyter的timeit获得正确的全局变量？

Question

第一个单元格我有这个：

from numba import cuda
@cuda.jit
def thread_counter_safe(global_counter):
    cuda.atomic.add(global_counter, 0, 1)  # Safely add 1 to offset 0 in global_counter array

在下一个单元格上，我有这个：

global_counter = cuda.to_device(np.array([0], dtype=np.int32))
thread_counter_safe[64, 64](global_counter)
print('Should be %d:' % (64*64), global_counter.copy_to_host())

global_counter = cuda.to_device(np.array([0], dtype=np.int32))
%timeit thread_counter_safe[64, 64](global_counter)
print('Should be %d:' % (64*64), global_counter.copy_to_host())

第二个单元格的输出如下：

Should be 4096: [4096]
10000 loops, best of 3: 118 µs per loop
Should be 4096: [168390656]

Jupyter Notebook的timeit在其迭代测试中带有global_counter。如何正确地回馈global_counter？