Why does changing a kernel parameter deplete my resources?

Question

I made a very simple kernel below to practice CUDA.

import pycuda.driver as cuda
import pycuda.autoinit
import numpy as np
from pycuda.compiler import SourceModule
from pycuda import gpuarray
import cv2

def compile_kernel(kernel_code, kernel_name):
  mod = SourceModule(kernel_code)
  func = mod.get_function(kernel_name)
  return func

input_file = np.array(cv2.imread('clouds.jpg'))
height, width, channels = np.int32(input_file.shape)

my_kernel_code = """
  __global__ void my_kernel(int width, int height) {
    // This kernel trivially does nothing! Hurray!
  }
"""
kernel = compile_kernel(my_kernel_code, 'my_kernel')

if __name__ == '__main__':

  for i in range(0, 2):
    print 'o'
    kernel(width, height, block=(32, 32, 1), grid=(125, 71))

    # When I take this line away, the error goes bye bye.
    # What in the world?
    width -= 1

Right now, if we run the code above, execution proceeds through the first iteration of the for loop just fine. However, during the second iteration of the loop, I get the following error.

Traceback (most recent call last):
  File "outOfResources.py", line 27, in <module>
    kernel(width, height, block=(32, 32, 1), grid=(125, 71))
  File "/software/linux/x86_64/epd-7.3-1-pycuda/lib/python2.7/site-packages/pycuda-2012.1-py2.7-linux-x86_64.egg/pycuda/driver.py", line 374, in function_call
    func._launch_kernel(grid, block, arg_buf, shared, None)
pycuda._driver.LaunchError: cuLaunchKernel failed: launch out of resources

If I take away the line width -= 1, the error goes away. Why is that? Can't I change the parameter for a kernel the second time around? For reference, here is clouds.jpg.

Answer 1

Though the error message isn't particularly informative, note that you need to pass in a correctly casted width variable. So something like:

width = np.int32(width - 1)

should work.