我可以使用生成的 swig 代码将 C++ 对象转换为 PyObject 吗?

问题描述

我正在使用 swig 将 python 嵌入到我的 C++ 程序中.目前我有一个用 C++ 编写的对象，我想将其传递给 python 函数.我已经创建了 swig 接口来包装类.

I'm working on embedding python into my C++ program using swig. At the moment I have a object written in C++ which I want to pass to a python function. I've created the swig interface to wrap the class.

我想要做的是将我创建的这个 C++ 对象传递给一个 python 函数，该函数可以像在 C++ 中一样使用它.我是否可以使用 swig 生成的代码来执行此操作?如果不是，我该如何处理?

What I'm trying to do is take this C++ object which I've created and pass it to a python function with the ability to use it like I would in C++. Is it possible for me to use code generate by swig to do this? If not how can I approach this?

推荐答案

您可以使用 PyObject_CallMethod 将新创建的对象传回给 python.假设 ModuleName.object 是一个 python 对象，它有一个名为 methodName 的方法，你想将一个新创建的 C++ 对象传递给你想要粗略的(从内存中，我不能立即测试)在 C++ 中执行此操作:

You can use PyObject_CallMethod to pass a newly created object back to python. Assuming ModuleName.object is a python object with a method called methodName that you want to pass a newly created C++ object to you want to roughly (from memory, I can't test it right now) do this in C++:

int callPython() {
   PyObject* module = PyImport_ImportModule("ModuleName");
   if (!module)
      return 0;

   // Get an object to call method on from ModuleName
   PyObject* python_object = PyObject_CallMethod(module, "object", "O", module);
   if (!python_object) {
      PyErr_Print();
      Py_DecRef(module);
      return 0;
   }

   // SWIGTYPE_p_Foo should be the SWIGTYPE for your wrapped class and
   // SWIG_POINTER_NEW is a flag indicating ownership of the new object
   PyObject *instance = SWIG_NewPointerObj(SWIG_as_voidptr(new Foo()), SWIGTYPE_p_Foo, SWIG_POINTER_NEW);

   PyObject *result = PyObject_CallMethod(python_object, "methodName", "O", instance);
   // Do something with result?

   Py_DecRef(instance);
   Py_DecRef(result);
   Py_DecRef(module);

   return 1;
}

我认为我有正确的引用计数，但我不完全确定.

I think I've got the reference counting right for this, but I'm not totally sure.

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