运行python numbapro时出现Cuda资源不足错误

我试图在numbapro python中运行一个cuda内核，但我一直遇到资源不足的错误。 然后，我尝试将内核执行到一个循环中，并发送较小的数组，但这仍然给了我相同的错误

以下是我的错误消息：

Traceback (most recent call last):
File "./predict.py", line 418, in <module>
predict[griddim, blockdim, stream](callResult_d, catCount, numWords, counts_d, indptr_d, indices_d, probtcArray_d, priorC_d)
File "/home/mhagen/Developer/anaconda/lib/python2.7/site-packages/numba/cuda/compiler.py", line 228, in __call__
sharedmem=self.sharedmem)
File "/home/mhagen/Developer/anaconda/lib/python2.7/site-packages/numba/cuda/compiler.py", line 268, in _kernel_call
cu_func(*args)
File "/home/mhagen/Developer/anaconda/lib/python2.7/site-packages/numba/cuda/cudadrv/driver.py", line 1044, in __call__
self.sharedmem, streamhandle, args)
File "/home/mhagen/Developer/anaconda/lib/python2.7/site-packages/numba/cuda/cudadrv/driver.py", line 1088, in launch_kernel
None)
File "/home/mhagen/Developer/anaconda/lib/python2.7/site-packages/numba/cuda/cudadrv/driver.py", line 215, in safe_cuda_api_call
self._check_error(fname, retcode)
File "/home/mhagen/Developer/anaconda/lib/python2.7/site-packages/numba/cuda/cudadrv/driver.py", line 245, in _check_error
raise CudaAPIError(retcode, msg)
numba.cuda.cudadrv.driver.CudaAPIError: Call to cuLaunchKernel results in CUDA_ERROR_LAUNCH_OUT_OF_RESOURCES

---

我发现这是在cuda程序中

调用predict时，blockdim设置为1024。 预测[griddim、blockdim、stream]（调用结果、catCount、NumWord、计数、索引、索引、ProbtArray、优先级）

但该过程是以1000个元素而不是1024个元素的切片大小进行迭代调用的。 因此，在该过程中，它将尝试写入返回数组中超出边界的24个元素

发送多个elements参数（n_el）并在cuda过程中放置错误检查调用可以解决此问题

@cuda.jit(argtypes=(double[:], int64, int64, int64, double[:], int64[:], int64[:], double[:,:], double[:] ))
def predict( callResult, n_el, catCount, wordCount, counts, indptr, indices, probtcArray, priorC ):

     i = cuda.threadIdx.x + cuda.blockIdx.x * cuda.blockDim.x

     if i < n_el:

         ....

@cuda.jit（argtypes=（双[：]，int64，int64，int64，双[：]，int64[：]，int64[：]，int64[：]，int64[：]，双[：]，双[：]））
def predict（调用结果、n_el、catCount、wordCount、counts、indptr、索引、probtcArray、priorC）：
i=cuda.threadIdx.x+cuda.blockIdx.x*cuda.blockDim.x
如果i

这不应该是向该部分产品的供应商提交的错误报告吗？

@cuda.jit(argtypes=(double[:], int64, int64, int64, double[:], int64[:], int64[:], double[:,:], double[:] ))
def predict( callResult, n_el, catCount, wordCount, counts, indptr, indices, probtcArray, priorC ):

     i = cuda.threadIdx.x + cuda.blockIdx.x * cuda.blockDim.x

     if i < n_el:

         ....