How to profile PyCuda code with the Visual Profiler?
When I create a new session and tell the Visual Profiler to launch my python/pycuda scripts I get following error message: Execution run #1 of program '' failed, exit code: 255
These are my preferences:
- Launch:
python "/pathtopycudafile/mysuperkernel.py"
- Working Directory:
"/pathtopycudafile/mysuperkernel.py"
- Arguments:
[empty]
I use CUDA 4.0 under Ubuntu 10.10. 64Bit. Profiling compiled examples works.
p.s. I am aware of SO question How to profile PyCuda code in Linux?, but seems to be an unrelated problem.
Minimal example
pycudaexample.py:
import pycuda.autoinit
import pycuda.driver as drv
import numpy
from pycuda.compiler import SourceModule
mod = SourceModule("""
__global__ void multiply_them(float *dest, float *a, float *b)
{
const int i = threadIdx.x;
dest[i] = a[i] * b[i];
}
""")
multiply_them = mod.get_function("multiply_them")
a = numpy.random.randn(400).astype(numpy.float32)
b = numpy.random.randn(400).astype(numpy.float32)
dest = numpy.zeros_like(a)
multiply_them(
drv.Out(dest), drv.In(a), drv.In(b),
block=(400,1,1), grid=(1,1))
pycuda.autoinit.context.detach()
Example settings
Error message
There is something wrong with the way you are specifying the executable to the compute profiler. If I put a hash bang line at the top of your posted code:
#!/usr/bin/env python
and then give the python file executable permissions, the compute profiler runs the code without complaint and I get this:
There are two methods that you can use.
Launch the Script Interpreter
Launch python
Arguments "/pathtopycudafile/mysuperkernel.py"
Launch a Executable Script
Launch "/pathtopycudafile/mysuperkernel.py"
Arguments [blank]
mysuperkernel.py must be executable (chmod +x)
mysuperkenrel.py must have a #! to specify the path to the interpreter
See @talonmies answer