在Python C扩展中处理PyList_Append时丢失在Py_DECREF/INCREF上

Question

处理PyList_Append时,我在Py_DECREF/INCREF上丢失了.任何人都可以对以下代码发表评论吗？

PyObject * bugmaybe(PyObject *self, PyObject *args)
{
   PyObject * trio=PyList_New(0);
   PyObject * trio_tmp;
   PyObject * otmp = PyFloat_FromDouble(1.2);
   PyList_Append(trio_tmp,otmp);
   //Py_DECREF(otmp);
   otmp = PyFloat_FromDouble(2.3);
   PyList_Append(trio_tmp,otmp);
   //Py_DECREF(otmp);
   PyList_Append(trio,trio_tmp);
   Py_INCREF(trio_tmp);
}

Answer 1

如果您事先知道列表的大小,通常可以更快地创建具有正确大小和使用的列表PyList_SetItem().

您的代码完全错误,trio_tmp未初始化.

试试这个:

PyObject * bugmaybe(PyObject *self, PyObject *args)
{
  PyObject * trio=PyList_New(3);
  PyObject * otmp = PyFloat_FromDouble(1.2);
  PyList_SetItem(trio,0,otmp);
  otmp = PyFloat_FromDouble(2.3);
  PyList_SetItem(trio,1,otmp);
  PyList_Append(trio,2, PyList_New(0));
  return trio;
}

如果你真的想要使用PyList_Append,你的代码基本上没问题,只是错过了初始化trio_tmp和最后多余的Py_INCREF.

PyObject * bugmaybe(PyObject *self, PyObject *args)
{
  PyObject * trio=PyList_New(0);
  // trio has refcount 1
  PyObject * trio_tmp = PyList_New(0);
  // trio_tmp has recount 1
  PyObject * otmp = PyFloat_FromDouble(1.2);
  // otmp has recount 1
  PyList_Append(trio_tmp,otmp);
  // Append does not steal a reference, so otmp refcoun = 2
  Py_DECREF(otmp);
  // otmp refcount = 1, but stored in the list so the pointer var
  // can be reused
  otmp = PyFloat_FromDouble(2.3);
  PyList_Append(trio_tmp,otmp);
  Py_DECREF(otmp);
  // as above
  PyList_Append(trio,trio_tmp);
  // decrement refcount for trio_tmp, as it has recount 2 now.
  Py_DECREF(trio_tmp);
  return trio;
}

上面的代码相当于:

 trio = []
 trio_tmp = []
 otmp = 1.2
 trio_tmp.append(otmp)
 otmp = 2.3
 trio_tmp.append(otmp)
 trio.append(trio_tmp)

希望能帮助到你.主要提示是在文档中,如果它说"窃取引用",那么函数基本上取得所有权,如果它说"新引用"然后它为你做了一个INCREF,如果没有说它可能会做一个INCREF和DECREF对如所须.