我想创建一个嵌入式python 3解释器,让python脚本导入通过C Python API创建的模块。
创建"顶级"没问题模块,但现在我想用软件包组织我的模块......但是我失败了。
这是我目前的(简单)代码:
#include <Python.h>
//// Definition of 'emb.sub' module
static PyObject* emb_sub_foo(PyObject *self, PyObject *args)
{
char const* n = "I am sub foo";
return Py_BuildValue("s", n);
}
static PyMethodDef EmbSubMethods[] = {
{"foo", emb_sub_foo, METH_VARARGS, "Returns sub foo"},
{NULL, NULL, 0, NULL}
};
static PyModuleDef EmbSubModule = {
PyModuleDef_HEAD_INIT, "emb.sub", NULL, -1, EmbSubMethods,
NULL, NULL, NULL, NULL
};
static PyObject* PyInit_emb_sub(void)
{
return PyModule_Create(&EmbSubModule);
}
//// Embedded Python
int main()
{
PyImport_AppendInittab("emb.emb", &PyInit_emb_sub);
Py_Initialize();
PyRun_SimpleString("import emb.sub\n");
Py_Finalize();
return 0;
}
当我执行程序时,我得到:
Traceback (most recent call last):
File "<string>", line 1, in <module>
ImportError: No module named 'emb'
所以我创建了一个空的emb模块,并将其__path__
设置为:
#include <Python.h>
//// Definition of 'emb' module
static PyModuleDef EmbModule = {
PyModuleDef_HEAD_INIT, "emb", NULL, -1, NULL,
NULL, NULL, NULL, NULL
};
static PyObject* PyInit_emb(void)
{
PyObject *mod = PyModule_Create(&EmbModule);
PyModule_AddObject(mod, "__path__", Py_BuildValue("()"));
return mod;
}
//// Definition of 'emb.sub' module
static PyObject* emb_sub_foo(PyObject *self, PyObject *args)
{
char const* n = "I am sub foo";
return Py_BuildValue("s", n);
}
static PyMethodDef EmbSubMethods[] = {
{"foo", emb_sub_foo, METH_VARARGS, "Returns sub foo"},
{NULL, NULL, 0, NULL}
};
static PyModuleDef EmbSubModule = {
PyModuleDef_HEAD_INIT, "emb.sub", NULL, -1, EmbSubMethods,
NULL, NULL, NULL, NULL
};
static PyObject* PyInit_emb_sub(void)
{
return PyModule_Create(&EmbSubModule);
}
//// Embedded Python
int main()
{
PyImport_AppendInittab("emb", &PyInit_emb);
PyImport_AppendInittab("emb.sub", &PyInit_emb_sub);
Py_Initialize();
PyRun_SimpleString("import emb.sub\n");
Py_Finalize();
return 0;
}
现在我收到了这个错误:
Traceback (most recent call last):
File "<string>", line 1, in <module>
ImportError: No module named 'emb.sub'
我想知道是否可以在嵌入式python中创建包和模块的层次结构?
谢谢!
答案 0 :(得分:3)
据我所知,默认的内置模块处理无法与包+模块名称匹配。这可以通过添加一个自定义导入钩子来解决,只需在Py_Initialize()之后运行它:
PyRun_SimpleString(
"import importlib.abc\n" \
"import importlib.machinery\n" \
"import sys\n" \
"\n" \
"\n" \
"class Finder(importlib.abc.MetaPathFinder):\n" \
" def find_spec(self, fullname, path, target=None):\n" \
" if fullname in sys.builtin_module_names:\n" \
" return importlib.machinery.ModuleSpec(\n" \
" fullname,\n" \
" importlib.machinery.BuiltinImporter,\n" \
" )\n" \
"\n" \
"\n" \
"sys.meta_path.append(Finder())\n" \
);
我用完整的代码尝试了这一点并取得了成功:
#include <Python.h>
//// Definition of 'emb' module
static PyModuleDef EmbModule = {
PyModuleDef_HEAD_INIT, "emb", NULL, -1, NULL,
NULL, NULL, NULL, NULL
};
static PyObject* PyInit_emb(void)
{
PyObject *mod = PyModule_Create(&EmbModule);
PyModule_AddObject(mod, "__path__", Py_BuildValue("()"));
return mod;
}
//// Definition of 'emb.sub' module
static PyObject* emb_sub_foo(PyObject *self, PyObject *args)
{
char const* n = "I am sub foo";
return Py_BuildValue("s", n);
}
static PyMethodDef EmbSubMethods[] = {
{"foo", emb_sub_foo, METH_VARARGS, "Returns sub foo"},
{NULL, NULL, 0, NULL}
};
static PyModuleDef EmbSubModule = {
PyModuleDef_HEAD_INIT, "emb.sub", NULL, -1, EmbSubMethods,
NULL, NULL, NULL, NULL
};
static PyObject* PyInit_emb_sub(void)
{
return PyModule_Create(&EmbSubModule);
}
//// Embedded Python
int main()
{
PyImport_AppendInittab("emb", &PyInit_emb);
PyImport_AppendInittab("emb.sub", &PyInit_emb_sub);
Py_Initialize();
PyRun_SimpleString(
"import importlib.abc\n" \
"import importlib.machinery\n" \
"import sys\n" \
"\n" \
"\n" \
"class Finder(importlib.abc.MetaPathFinder):\n" \
" def find_spec(self, fullname, path, target=None):\n" \
" if fullname in sys.builtin_module_names:\n" \
" return importlib.machinery.ModuleSpec(\n" \
" fullname,\n" \
" importlib.machinery.BuiltinImporter,\n" \
" )\n" \
"\n" \
"\n" \
"sys.meta_path.append(Finder())\n" \
);
PyRun_SimpleString("import emb.sub\n");
PyRun_SimpleString("print(emb.sub.foo())\n");
Py_Finalize();
return 0;
}