[lkml]   [2008]   [Jan]   [16]   [last100]   RSS Feed
Views: [wrap][no wrap]   [headers]  [forward] 
Messages in this thread
    SubjectRe: Why is the kfree() argument const?

    On Wed, 16 Jan 2008, Johannes Weiner wrote:
    > is there any reason why kfree() takes a const pointer just to degrade it
    > with the call to slab_free()/__cache_free() again? The promise that the
    > pointee is not modified is just bogus in this case, anyway, isn't it?

    "const" has *never* been about the thing not being modified. Forget all
    that claptrap. C does not have such a notion.

    "const" is a pointer type issue, and is meant to make certain mis-uses
    more visible at compile time. It has *no* other meaning, and anybody who
    thinks it has is just setting himself up for problems.

    In the particular case of kfree(), the pointer argument *should* be const,
    and the fact that the C library gets this wrong is not the kernels
    problem, it's a problem with the C library.


    - From a very obvious and very *real* caller perspective, "free()" really
    doesn't change the thing the pointer points to. It does something
    totally different: it makes the *pointer* itself invalid.

    In other words, if you think that "kfree()" changed the thing you
    free'd, you're simply wrong. It did no such thing. The memory is 100%
    the same, it's just that you cannot access it any more, and if you try,
    you'll get somebody elses memory.

    In other words, "kfree()" can be const.

    - Anything that *can* take a const pointer should always do so.

    Why? Because we want the types to be as tight as possible, and normal
    code should need as few casts as possible.

    Here's a question for you: let's say that you have a structure that
    has a member that is never changed. To make that obvious, and to allow
    the compiler to warn about mis-use of a pointer, the structure should
    look something like

    struct mystruct {
    const char *name;

    and let's look at what happens if the allocation of that const thing is

    The *correct* way to do that is:

    char *name = kmalloc(...)
    /* Fill it in */
    snprintf(name, ...)
    mystruct->name = name;

    and there are no casts anywhere, and you get exactly the semantics you
    want: "name" itself isn't constant (it's obviously modified), but at
    the same time the type system makes it very clear that trying to change
    it through that mystruct member pointer is wrong.

    How do you free it?

    That's right, you do:


    and this is why "kfree()" should take a const pointer. If it doesn't,
    you have to add an *incorrect* and totally useless cast to code that
    was correct.

    So never believe that "const" is some guarantee that the memory under the
    pointer doesn't change. That is *never* true. It has never been true in
    C, since there can be arbitrary pointer aliases to that memory that aren't
    actually const. If you think "const *p" means that the memory behind "p"
    is immutable, you're simply wrong.

    Anybody who thinks that kfree() cannot (or should not) be const doesn't
    understand the C type system.


     \ /
      Last update: 2008-01-16 19:41    [W:0.022 / U:24.872 seconds]
    ©2003-2017 Jasper Spaans. hosted at Digital OceanAdvertise on this site