在 Python SWIG 中包装一个 void * 参数 [英] Wrapping a void * argument in Python SWIG

问题描述

我正在用 Python SWIG 包装这个库，它的函数如下所示:

I am wrapping this lib with Python SWIG that has a function looking like this:

int set_option(Foo *foo, const char *name, void *value);

在库中 const char *name 映射到我可以访问的类型:int, char *, <代码>字符**.

In the lib const char *name is mapped to a type that I have access to look up: int, char *, char **.

默认生成的包装器代码只接受一个包装好的void *(自然).

The wrapper code generated by default accepts only a wrapped void * (naturally).

让包装方法接受任何 Python 对象作为参数，并在我自己的 C 代码中进行类型检查和 Python 到 C 的转换的最佳方法是什么?

What is the best way to make the wrapped method accept any Python object as argument, and do the type checking, and Python to C conversion in my own C code?

我的猜测是某种类型的地图，但可惜我无法弄清楚.

My guess would be some kind of type map, but alas I cannot figure it out.

@Goodies

In [12]: lib.set_option(foo, "option", "value")
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-12-4ccf706b5c50> in <module>()
----> 1 lib.set_option(foo, "option", "value")

TypeError: in method 'set_option', argument 3 of type 'void *'

推荐答案

如果您只想将 PyObject 传递给使用 %extend 的内容，您可以修改您的内容已经得到并简单地写下以下内容:

If you just want to pass a PyObject to something using %extend you can modify what you've got and simply write the following:

%extend Foo {
    int set_option(const char *name, PyObject *value) {
            // Here value is now a PyObject *, so work with that.
            // $self is Foo *foo
            return 0;
    }
};

不需要类型映射，这个函数现在可以接受任何 Python 输入.你用它做什么取决于你，如果你保留对象，你需要增加引用计数.

There's no need for a typemap and this function now accepts any Python input. What you do with it is up to you and if you retain the object you'll need increase the reference count.

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