如何使用纯 Python 扩展 API (python3) 包装 C++ 对象？

Question

我想知道如何在没有外部工具（如 Cython、Boost、SWIG 等）的情况下使用 Python Extension API（和 distutils）包装 C++ 对象。只是以纯粹的 Python 方式，无需创建 dll。

请注意，我的 C++ 对象有内存分配，因此必须调用析构函数以避免内存泄漏。

#include "Voice.h"

namespace transformation
{ 
  Voice::Voice(int fftSize) { mem=new double[fftSize]; } 

  Voice::~Voice() { delete [] mem; } 

  int Voice::method1() { /*do stuff*/ return (1); } 
}

我只想在 Python 中做类似的事情：

import voice

v=voice.Voice(512)
result=v.method1()

Answer 1

似乎答案实际上就在这里：https://docs.python.org/3.6/extending/newtypes.html

有例子，但并不容易。

编辑 1：

事实上，它并不是真正将C++对象包装在Python对象中，而是用C代码创建一个Python对象。（edit2：这样你就可以包装 C++ 对象了！）

编辑 2：

这是一个使用 Python newtypes

的解决方案

.

原始 C++ 文件：Voice.cpp

#include <cstdio>

#include "Voice.h"

namespace transformation
{ 
    Voice::Voice(int fftSize) {
        printf("c++ constructor of voice\n");
        this->fftSize=fftSize;
        mem=new double[fftSize];
        } 

    Voice::~Voice() { delete [] mem; } 

    int Voice::filter(int freq) {
        printf("c++ voice filter method\n");
        return (doubleIt(3));
    }
    int Voice::doubleIt(int i) { return 2*i; }
}

.

原始 h 文件：Voice.h

namespace transformation {

    class Voice {
    public:
        double *mem;
        int fftSize;

        Voice(int fftSize);
        ~Voice();

        int filter(int freq);
        int doubleIt(int i);
    };

}

.

C++ Python 包装文件：voiceWrapper.cpp

#include <Python.h>

#include <cstdio>
//~ #include "structmember.h"

#include "Voice.h"

using transformation::Voice;

typedef struct {
    PyObject_HEAD
    Voice * ptrObj;
} PyVoice;




static PyModuleDef voicemodule = {
    PyModuleDef_HEAD_INIT,
    "voice",
    "Example module that wrapped a C++ object",
    -1,
    NULL, NULL, NULL, NULL, NULL
};

static int PyVoice_init(PyVoice *self, PyObject *args, PyObject *kwds)
// initialize PyVoice Object
{
    int fftSize;

    if (! PyArg_ParseTuple(args, "i", &fftSize))
        return -1;

    self->ptrObj=new Voice(fftSize);

    return 0;
}

static void PyVoice_dealloc(PyVoice * self)
// destruct the object
{
    delete self->ptrObj;
    Py_TYPE(self)->tp_free(self);
}


static PyObject * PyVoice_filter(PyVoice* self, PyObject* args)
{
    int freq;
    int retval;

    if (! PyArg_ParseTuple(args, "i", &freq))
        return Py_False;

    retval = (self->ptrObj)->filter(freq);

    return Py_BuildValue("i",retval);
}


static PyMethodDef PyVoice_methods[] = {
    { "filter", (PyCFunction)PyVoice_filter,    METH_VARARGS,       "filter the mem voice" },
    {NULL}  /* Sentinel */
};

static PyTypeObject PyVoiceType = { PyVarObject_HEAD_INIT(NULL, 0)
                                    "voice.Voice"   /* tp_name */
                                };


PyMODINIT_FUNC PyInit_voice(void)
// create the module
{
    PyObject* m;

    PyVoiceType.tp_new = PyType_GenericNew;
    PyVoiceType.tp_basicsize=sizeof(PyVoice);
    PyVoiceType.tp_dealloc=(destructor) PyVoice_dealloc;
    PyVoiceType.tp_flags=Py_TPFLAGS_DEFAULT;
    PyVoiceType.tp_doc="Voice objects";
    PyVoiceType.tp_methods=PyVoice_methods;
    //~ PyVoiceType.tp_members=Noddy_members;
    PyVoiceType.tp_init=(initproc)PyVoice_init;

    if (PyType_Ready(&PyVoiceType) < 0)
        return NULL;

    m = PyModule_Create(&voicemodule);
    if (m == NULL)
        return NULL;

    Py_INCREF(&PyVoiceType);
    PyModule_AddObject(m, "Voice", (PyObject *)&PyVoiceType); // Add Voice object to the module
    return m;
}

.

distutils 文件：setup.py

from distutils.core import setup, Extension

setup(name='voicePkg', version='1.0',  \
      ext_modules=[Extension('voice', ['voiceWrapper.cpp','Voice.cpp'])])

.

python 测试文件：test.py

import voice

v=voice.Voice(512)
result=v.filter(5)
print('result='+str(result))

.

和魔法：

sudo python3 setup.py install
python3 test.py

输出为：

c++ constructor of voice
c++ voice filter method
result=6

尽情享受吧！

厄运

如何使用纯 Python 扩展 API (python3) 包装 C++ 对象？

How to wrap a C++ object using pure Python Extension API (python3)?

c++

python

word-wrap

python-3.x