New submission from Sebastian Berg <sebast...@sipsolutions.net>:
The current documentation of ``PyBuffer_Release()`` and the PEP is a bit fuzzy
about what the function can and cannot do.
When an object exposes the buffer interface, I believe it should always return
a `view` (in NumPy speak) of its own data, i.e. the data exposed by the object
is also owned by it directly.
On the other hand the buffer view _itself_ has fields such as `strides`, etc.
which may need allocating.
In other words, I think `PyBuffer_Release()` should be documented to
deallocate/invalidate the `Py_buffer`. But, it *must not* invalidate the actual
memory it points to. If I copy all information out of the `Py_buffer` and then
free it, the copy must still be valid.
I think this is the intention, but it is not spelled out clear enough, it is
also the reason for the behaviour of the "#s", etc. keyword argument parsers
failing due to the code:
if (pb != NULL && pb->bf_releasebuffer != NULL) {
*errmsg = "read-only bytes-like object";
return -1;
}
which in turn currently means NumPy decides to _not_ implement bf_releasebuffer
at all (leading to very ugly work arounds).
I am happy to make a PR, if we can get to a point where everyone is absolutely
certain that the above interpretation was always correct, we could clean up a
lot of code inside NumPy as well!
----------
components: C API
messages: 360809
nosy: seberg
priority: normal
severity: normal
status: open
title: Meaning and clarification of PyBuffer_Release()
type: behavior
versions: Python 2.7, Python 3.5, Python 3.6, Python 3.7, Python 3.8, Python 3.9
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Python tracker <rep...@bugs.python.org>
<https://bugs.python.org/issue39471>
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