How Python Implements "long integer"?
pm567426 at gmail.com
Sun Jul 5 18:32:41 CEST 2009
On Jul 5, 8:12 pm, a... at pythoncraft.com (Aahz) wrote:
> In article <6f6be2b9-49f4-4db0-9c21-52062d8ea... at l31g2000yqb.googlegroups.com>,
> Pedram <pm567... at gmail.com> wrote:
> >This time I have a simple C question!
> >As you know, _PyLong_New returns the result of PyObject_NEW_VAR. I
> >found PyObject_NEW_VAR in objimpl.h header file. But I can't
> >understand the last line :( Here's the code:
> >#define PyObject_NEW_VAR(type, typeobj, n) \
> >( (type *) PyObject_InitVar( \
> > (PyVarObject *) PyObject_MALLOC(_PyObject_VAR_SIZE((typeobj),
> >(n)) ),\
> > (typeobj), (n)) )
> >I know this will replace the PyObject_New_VAR(type, typeobj, n)
> >everywhere in the code and but I can't understand the last line, which
> >is just 'typeobj' and 'n'! What do they do? Are they make any sense in
> >allocation process?
> Look in the code to find out what PyObject_InitVar() does -- and, more
> importantly, what its signature is. The clue you're missing is the
> trailing backslash on the third line, but that should not be required if
> you're using an editor that shows you matching parentheses.
> Aahz (a... at pythoncraft.com) <*> http://www.pythoncraft.com/
> "as long as we like the same operating system, things are cool." --piranha
No, they wrapped the 3rd line!
I'll show you the code in picture below:
As you can see the PyObject_MALLOC has nothing to do with typeobj and
n in line 4.
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