On 05/03/2026 19:44, Andrew Davis wrote:

On 3/5/26 9:59 AM, Krzysztof Kozlowski wrote:

quoted

On 05/03/2026 16:52, Andrew Davis wrote:

quoted

quoted

quoted

quoted

The code is not correct logically, either, because functions like
ti_sci_get_handle() and ti_sci_put_handle() are meant to modify the
handle reference counting, thus they must modify the handle.

The reference counting is handled outside of the ti_sci_handle struct,
the contents of the handle are never modified after it is created.

The const is only added by functions return a handle to consumers.
We cannot return non-const to consumer drivers or then they would
be able to modify the content without a compiler warning, which would
be a real problem.

This is the same argument as making pointer to const the pointer freed
via kfree() (or free() in userspace). kfree() does not modify the
contents of the pointer, right? The same as getting putting handle does
not modify the handle...

In that argument, if we wanted the consumer of the pointer to not free()
it we would return a const pointer, free()'ing that would result in the
warning we want (discards const qualifier).

If you could somehow malloc() from a const area in memory then free()
doesn't modify the pointed to values, only the non-const record keeping
which would be stored outside of the const memory. So even in this analogy
there isn't a problem.

I am not saying about malloc. I am saying about free() which does not
modify the freed memory.

And if you look, kfree() in Linux takes a const pointer. We also do not

The slub, but that's the only implementation being I believe frowned
upon. The mistake made long time ago...

modify the content of the pointer we are given either, so we should
be okay using const by the same reasoning.

That's a mistake so you cannot use the same reasoning. It's bogus and
bugfree to take a pointer to const for any kfree(). Just poke MM folks...

quoted

quoted

quoted

The point is that storing the reference counter outside of handle does
not make the argument correct. Logically when you get a reference, you
increase the counter, so it is not a pointer to const. And the code
agrees, because you must drop the const.

The record keeping memory is not const and can be modified.

And where do we drop the const? The outer "struct ti_sci_info" was never
const to begin with, so no dropped const.

We discuss about different points. I did not say the outer memory is
const. I said that you drop the const - EXPLICITLY - from the pointer to
handle.

Only because container_of() forces the const to be dropped, that is out
of our control. But we never modify handle though the non-const parent
struct.

That is not true. You could use container_of_const() if you wanted to
have const. You explicitly drop the const, code would not work without
dropping the const and this is the problem.

quoted

And that API which gets a handle (increases reference count) via pointer
to const is completely illogical, because increasing refcnt is already
modifying it. Just because you store the refcnt outside, does not change
the fact that API is simply confusing.

If the refcnt is not inside the const struct, then the contents are not
changed, therefor const is still correct. Even if the content of handle
were in fixed ROM, nothing would break here.

I am talking about API and again you go into memory correctness. So
again, very simple: any refcnt get taking const data is bogus.


Best regards,
Krzysztof