Re: [PATCH v2] setup: don't die if realpath(3) fails on getcwd(3)

[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

 



On Fri, May 27, 2022 at 1:02 AM Ævar Arnfjörð Bjarmason
<avarab@xxxxxxxxx> wrote:
>
>
> On Tue, May 24 2022, Elijah Newren wrote:
>
> > [...] So, I think the signature of free() is just
> > wrong: it should have taken a const void* all along.  Unfortunately,
> > the wrong type signature sadly makes people feel like they have to
> > choose between (a) dropping the added safety of const that the
> > compiler can enforce for you during the lifetime of the variable, or
> > (b) leaking memory you no longer need.
>
> Hrm, don't you mean that it would be better as:
>
>         void free(void *const ptr);
>
> Not:
>
>         void free(const void *ptr);

Nope, I definitely meant the latter; the stuff pointed to is const,
not the pointer itself.

In fact, I don't see any point at all in the former; with the free()
that exists today:

    void free(void *ptr)

I can pass it a "void * const myptr" already without problems, because
free's ptr parameter will be a copy of myptr, and thus modifying ptr
cannot affect myptr.  So such a call signature change could not
possibly provide any benefit to the outside caller.  But that call
signature change could hinder the actual implementation of free() for
some folks (particularly if a given implementation of free() keeps
extra data near the allocated block with information about the size of
the block and the next allocated block in the list).

In contrast, I cannot pass a "const void *myptr" or "const char
*myptr" to free(), but only because of the current type signature;
free() doesn't actually modify any of the contents the pointer points
to.  (And if you want to claim that free effectively does modify what
myptr points to because someone else could allocate that same memory,
remember that use-after-free is undefined regardless of whether the
data pointed to is const or not, and thus you cannot access that data
after free with or without the const.)  So, free()'s real type that it
acts on is a const void *.  Sadly, the declared type signature is
rather void *, which unnecessarily forces users to cast their types
when calling.




[Index of Archives]     [Linux Kernel Development]     [Gcc Help]     [IETF Annouce]     [DCCP]     [Netdev]     [Networking]     [Security]     [V4L]     [Bugtraq]     [Yosemite]     [MIPS Linux]     [ARM Linux]     [Linux Security]     [Linux RAID]     [Linux SCSI]     [Fedora Users]

  Powered by Linux