Python 扩展 : symbol(s) not found for architecture x86_64 error

Question

我想用 c 写一个 python 扩展。我在 Mac 上工作，我从 here 中获取了代码:

#include <Python.h>

static PyObject* say_hello(PyObject* self, PyObject* args)
{
    const char* name;

    if (!PyArg_ParseTuple(args, "s", &name))
        return NULL;

    printf("Hello %s!\n", name);

    Py_RETURN_NONE;
}

static PyMethodDef HelloMethods[] =
{
     {"say_hello", say_hello, METH_VARARGS, "Greet somebody."},
     {NULL, NULL, 0, NULL}
};

PyMODINIT_FUNC

inithello(void)
{
     (void) Py_InitModule("hello", HelloMethods);
}

我编译它:

gcc -c -o py_module.o py_module.c -I/Library/Frameworks/Python.framework/Versions/2.7/include/python2.7/
gcc -o py_module py_module.o -I/Library/Frameworks/Python.framework/Versions/2.7/include/python2.7/ -lm

但是我得到这个错误:

Undefined symbols for architecture x86_64:
  "_PyArg_ParseTuple", referenced from:
      _say_hello in py_module.o
  "_Py_InitModule4_64", referenced from:
      _inithello in py_module.o
  "__Py_NoneStruct", referenced from:
      _say_hello in py_module.o
  "_main", referenced from:
      start in crt1.10.6.o
ld: symbol(s) not found for architecture x86_64
collect2: ld returned 1 exit status
make: *** [py_module] Error 1

为什么python不支持X86_64架构？

Answer 1

两件事:

• 您需要将您的扩展链接为一个共享对象(您正在尝试链接一个可执行文件，这就是链接器寻找 main() 的原因)；
• 您需要链接到 Python 静态库 (-lpython)。