It's in the code. See longobject.c:

Py_SIZE(v) = ndigits*sign;

You can also see Py_SIZE(v) used on PyLongs all over the place in longobject.c, for example:

    v = (PyLongObject *)vv;
    i = Py_SIZE(v);

Just do a ctrl-f for Py_SIZE(v) in longobject.c. Like I said, by looking in the implementation I was able to figure out that Py_SIZE is interpreted as the sign times the number of digits (unless I'm missing something), but this should be in the docs IMO.


On Wed, Oct 19, 2016 at 7:24 PM Thomas Nyberg <tomuxiong@gmx.com> wrote:
On 10/19/2016 09:04 PM, Elliot Gorokhovsky wrote:
> A quick note:
>
> I'm working on a special-case compare function for bounded integers for
> the sort stuff. By looking at the implementation, I figured out that
> Py_SIZE of a long is the sign times the number of digits (...right?).
> Before looking at the implementation, though, I had looked for this info
> in the docs, and I couldn't find it anywhere. Since Py_SIZE is public, I
> think the documentation should make clear what it returns for PyLongs,
> for example somewhere on the "Integer Objects" page. Apologies if this
> is specified somewhere else in the docs and I just couldn't find it.
>
> Elliot

I don't think this is right.

        https://github.com/python/cpython/blob/master/Include/object.h#L119
        https://docs.python.org/3/c-api/structures.html#c.Py_SIZE
        https://docs.python.org/3/c-api/structures.html#c.PyVarObject

It returns the `ob_size` fields of a PyVarObject. I think this has to do
with objects with variable sizes like lists. PyLongs are not
PyVarObjects because they have no notion of length.

Why would a long be stored as a sequence of digits instead of a (say) 64
bit integer as 8 bytes?

Cheers,
Thomas

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