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如何使用纯Python扩展API(python3)包装C ++对象?

如何使用纯Python扩展API(python3)包装C ++对象?

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

有例子,但并不是很容易。

编辑1:

实际上,并不是将C 对象包装在Python对象中,而是使用C 代码创建Python对象。(edit2:所以您可以包装C 对象!)

编辑2:

这是使用Python新类型解决方

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包装器文件: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))

和魔术:

sudo python3 setup.py install
python3 test.py

输出为:

c++ constructor of voice
c++ voice filter method
result=6
python 2022/1/1 18:51:20 有373人围观

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