Re: No warning at implicit removal of const. Was: relearning C: why does an in-place change to a char* segfault?

On 8/4/2024 6:06 PM, Ben Bacarisse wrote:

"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
 

On 8/2/2024 3:29 PM, Ben Bacarisse wrote:

"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>

For some reason I had a sort of a habit wrt const pointers:
>
(experimental code, no ads, raw text...)
https://pastebin.com/raw/f52a443b1
>
________________________________
/* Interfaces
____________________________________________________________________*/
#include <stddef.h>
>
>
struct object_prv_vtable {
    int (*fp_destroy) (void* const);
};
>
>
struct device_prv_vtable {
    int (*fp_read) (void* const, void*, size_t);
    int (*fp_write) (void* const, void const*, size_t);
};

Why?  It seems like an arbitrary choice to const qualify some pointer
types and some pointed-to types (but never both).

>
I just wanted to get the point across that the first parameter, aka, akin
to "this" in C++ is a const pointer. Shall not be modified in any way shape
or form. It is as it is, so to speak:
>
void foo(struct foobar const* const self);
>
constant pointer to a constant foobar, fair enough?

 No.  If you intended a const pointer to const object why didn't you
write that?  My point was that the consts seems to be scattered about
without any apparent logic and you've not explained why.
 

;^)

Does the wink mean I should not take what you write seriously?  If so,
please ignore my question.

>
The wink was meant to show my habit in basically a jestful sort of
way.

 Your habit of what?
 

To write the declaration with names and the const access I want, so:
extern void (void const* const ptr);
void (void const* const ptr)
{
    // ptr is a const pointer to a const void
}