指定* .pyd输出路径? [英] specify *.pyd output path?

问题描述

在此帖子中，答案是.c文件的输出路径.

In this post, the answer is for the output path of .c files.

我想知道如何将所有.pyd文件输出到特定目录(默认情况下，它与源代码混合)，例如，源代码中的目录pyd/.

I want to know how can I output all the .pyd file to a specific directory (by default, it mingles with the source code), for example, the directory pyd/ in the source code.

推荐答案

您可以将命令行参数--build-lib your/desired/output/path传递给执行setup.py的python调用.

You can pass the command-line argument --build-lib your/desired/output/path to the python invocation which executes setup.py.

此外，此选项与--inplace不兼容.如果您正在使用它，请将其删除.

Also, this option is not compatible with --inplace. If you're using that, remove it.

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