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From: Julio Di Egidio <julio@diegidio.name>
Newsgroups: comp.lang.c
Subject: Re: So You Think You Can Const?
Date: Sat, 11 Jan 2025 17:33:46 +0100
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On 11/01/2025 17:21, Julio Di Egidio wrote:
> On 11/01/2025 17:07, Julio Di Egidio wrote:
>> On 11/01/2025 12:14, David Brown wrote:
>>> On 10/01/2025 19:56, Keith Thompson wrote:
>> <snip>
>>> The idea was to place the emphasis on "free" changing the pointer, 
>>> rather than the data pointed to.
>>
>> I feel I am still altogether missing the point.
>>
>> Is my understanding correct that when freeing a pointer: 1) the 
>> pointer value, i.e. the address it holds, does not change; OTOH, 2) 
>> the pointed-to object does change, in the sense that it is marked 
>> unusable (and, supposedly, made available to re-allocation)?
>>
>> Moreover, while the pointer value has not changed, it is in fact 
>> changed in the sense that it has become invalid, namely the pointer 
>> cannot be used (validly dereferenced) anymore.  Not just that, but 
>> *every* pointer to the same object, i.e. holding the same address, has 
>> become invalid.
>>
>> All that considered, how isn't `void free(void *p)`, i.e. with no 
>> const qualifiers anywhere, the only reasonable signature?
> 
> In fact, along that line, I could see one might insist that "strictly 
> speaking, it should be `void free(void *const p)` because the pointer 
> value is not changed" (my considerations above are indeed more 
> "semantic"), OTOH, I just cannot see a case for `void free(void const 
> *p)`, not even strictly technically.

P.S.  Sorry, I might have snipped too much: to be clear, David Brown is 
actually referring to a `void free(void ** p)` in his statement above, 
so his "changing the pointer" is to be read in that context: I am back 
to the question of is there a problem with the standard signature of `free`.

-Julio