Python ctypes: loading DLL from from a relative path

Question

I have a Python module, wrapper.py, that wraps a C DLL. The DLL lies in the same folder as the module. Therefore, I use the following code to load it:

myDll = ctypes.CDLL("MyCDLL.dll")

This works if I execute wrapper.py from its own folder. If, however, I run it from elsewhere, it fails. That’s because ctypes computes the path relative to the current working directory.

My question is, is there a way by which I can specify the DLL’s path relative to the wrapper instead of the current working directory? That will enable me to ship the two together and allow the user to run/import the wrapper from anywhere.

Asked By: Frederick The Fool

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Source

Answer 1

You can use os.path.dirname(__file__) to get the directory where the Python source file is located.

Answered By: Matthew Flaschen

Answer 2

I always add the directory where my DLL is to the path. That works:

os.environ['PATH'] = os.path.dirname(__file__) + ';' + os.environ['PATH']
windll.LoadLibrary('mydll.dll')

Note that if you use py2exe, this doesn’t work (because __file__ isn’t set). In that case, you need to rely on the sys.executable attribute (full instructions at http://www.py2exe.org/index.cgi/WhereAmI)

Answered By: Chris B.

Answer 3

Expanding on Matthew’s answer:

import os.path
dll_name = "MyCDLL.dll"
dllabspath = os.path.dirname(os.path.abspath(__file__)) + os.path.sep + dll_name
myDll = ctypes.CDLL(dllabspath)

This will only work from a script, not the console nor from py2exe.

Answered By: fmark

Answer 4

Another version:

dll_file = os.path.join(os.path.dirname(os.path.abspath(__file__)), 'MyCDLL.dll')
myDll = ctypes.CDLL(dll_file)

Answered By: Shtefan

Python ctypes: loading DLL from from a relative path

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