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> Hmm, there's a misunderstanding. bf_releasebuffer is called exactly
> once for each call to bf_getbuffer.
Wrong: http://bugs.python.org/issue7433
static int
memory_getbuf(PyMemoryViewObject *self, Py_buffer *view, int flags)
{
 int res = 0;
 CHECK_RELEASED_INT(self);
 if (self->view.obj != NULL)
 res = PyObject_GetBuffer(self->view.obj, view, flags);
 if (view)
 dup_buffer(view, &self->view);
 return res;
}
After this, PyBuffer_Release will be called twice: once on the data in *view, by whoever acquired the buffer from memoryview, and once on self->view, by memory_dealloc. Both with the same bit-by-bit content of the Py_buffer structure.
Because there are two Py_buffer structures here, setting view.obj to NULL in PyBuffer_Release does not guarantee correct calls to bf_releasebuffer.
Note that the view.internal pointer is also clobbered above.
> > So, `bf_releasebuffer` cannot rely on (i) the data in Py_buffer
> > being what `bf_getbuffer` put there,
> 
> Well, why should it rely on that?
Because that makes implementing the exporter much easier. Also, writing an implementation for MemoryViewObject does not require clobbering the structure, and I doubt it helps much.
> > So, `bf_releasebuffer` cannot be used to release any resources
> > allocated in `bf_getbuffer`.
>
> AFAICT, it can. That's what the "internal" pointer is for.
Sure, guaranteeing that view->internal pointer is not toyed with would also be enough.
But the documentation should spell out very explicitly what the bf_releasebuffer call can rely on.