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Home/ Questions/Q 8812311
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Editorial Team
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Editorial Team
Asked: June 14, 20262026-06-14T03:34:35+00:00 2026-06-14T03:34:35+00:00

I am trying to teach myself CUDA. This has not been easy so far,

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I am trying to teach myself CUDA. This has not been easy so far, but I don’t give up easily either 🙂

I have created a very simple program. It merely returns a value from the GPU.

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

returnValue = np.zeros(1)
mod = SourceModule("""
  __global__ void myVeryFirstKernel(float* returnValue) {
    returnValue[0] = 8.0;
  }
""")
func = mod.get_function('myVeryFirstKernel')
func(cuda.InOut(returnValue), block=(1024, 1, 1), grid=(1, 1))
print str(returnValue[0])

However, the value that my program prints is 5.387879938e-315. That sure doesn’t look like 8.0. Why is the wrong value getting returned from the GPU?

I have tried altering the block size, which I don’t think should anything (but who knows). I have also checked that the data type I am sending in (float64) matches my kernel.

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  1. Editorial Team
    Editorial Team
    2026-06-14T03:34:37+00:00Added an answer on June 14, 2026 at 3:34 am

    You have type conflicts – your kernel is expected a 32 bit single precision value, but you are passing a 64 bit double value to it. If you rewrite your code something like this:

    returnValue = np.zeros(1, dtype=np.float32)
    mod = SourceModule("""
      __global__ void myVeryFirstKernel(float* returnValue) {
        returnValue[0] = 8.0f;
      }
    """)
    func = mod.get_function('myVeryFirstKernel')
    func(cuda.InOut(returnValue), block=(1024, 1, 1), grid=(1, 1))
    print returnValue[0]
    

    so that everything is explicitly specified in single precision, you might have more luck.

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