Python ctypes：从相对路径加载DLL [英] Python ctypes: loading DLL from from a relative path

问题描述

我有一个Python模块 wrapper.py ，该模块包装了一个C DLL。 DLL位于与模块相同的文件夹中。因此，我使用以下代码加载它：

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")

如果我从自己执行 wrapper.py 夹。但是，如果我从其他地方运行它，它将失败。那是因为ctypes计算相对于当前工作目录的路径。

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.

我的问题是，有没有一种方法可以指定DLL相对于包装器的路径而不是当前的工作目录？这样一来，我就可以将两者一起运送，并允许用户从任何地方运行/导入包装器。

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.

推荐答案

os.path.dirname（__ file __）获取Python源文件所在的目录。

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

这篇关于Python ctypes：从相对路径加载DLL的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！