Re: Why is the kfree() argument const?

[Chen Gang F T]

Hello Antoine:

  after read through the whole reply of Linus Torvalds for it
    (the time stamp is "Wed, 16 Jan 2008 10:39:00 -0800 (PST)").

  at least for me, his reply is correct in details.

  although what you said is also correct,
  it seems you misunderstanding what he said.

  all together:
    kfree() should use 'const void *' as parameter type
    the free() of C Library is incorrect (it use void *).


ä 2013å01æ13æ 03:18, antoine.trux@xxxxxxxxx åé:
> On Wednesday, January 16, 2008 8:39:48 PM UTC+2, Linus Torvalds wrote:
> 
>> "const" has *never* been about the thing not being modified. Forget all 
>> that claptrap. C does not have such a notion.
> 
> I beg your pardon?!
> 
> C has had that very notion ever since its first standard (1989). Here is an excerpt from that standard (ISO/IEC 9899:1990, section 6.5.3):
> 
>     "If an attempt is made to modify an object defined with a const-qualified type through use of an lvalue with non-const-qualified type, the behavior is undefined."
> 
> 
>> "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.
> 
> 'const' is also a pointer issue, but not only - see above quote from the C Standard.
> 
> 
> Defining an object 'const' can have an impact on optimization (and also on whether the object is placed in read-only memory). Here are trivial examples to illustrate:
> 
> <Program1>
> 
>     <foo1.c>
>     void foo1(const int* pi)
>     {
>         *(int*)pi = 1;
>     }
>     </foo1.c>
> 
>     <main1.c>
>     #include <stdio.h>
>     void foo1(const int* pi);
>     int main(void)
>     {
>         int i = 0;
>         foo1(&i);
>         printf("i = %d\n", i);
>         return 0;
>     }
>     </main1.c>
> 
> </Program1>
> 
> Program1 defines 'i' non-const, and modifies it through a const pointer, by casting const away in foo1(). This is allowed - although not necessarily wise.
> 
> Program1 has well defined behavior: it prints "i = 1". The generated code dutifully retrieves the value of 'i' before passing it to printf().
> 
> 
> <Program2>
> 
>     <foo2.c>
>     void foo2(const int* pi)
>     {
>     }
>     </foo2.c>
> 
>     <main2.c>
>     #include <stdio.h>
>     void foo2(const int* pi);
>     int main(void)
>     {
>         const int i = 0;
>         foo2(&i);
>         printf("i = %d\n", i);
>         return 0;
>     }
>     </main2.c>
> 
> </Program2>
> 
> Program2 defines 'i' const. A pointer to 'i' is passed to foo2(), which does not modify 'i'.
> 
> Program2 has well defined behavior: it prints "i = 0". When it generates code for main1.c, the compiler can assume that 'i' is not modified, because 'i' is defined const.
> 
> When compiling main2.c with gcc 4.4.7 with optimizations turned off (-O0), the generated code retrieves the value of 'i' before passing it to printf(). With optimizations turned on (-O3), it inlines the value of 'i', 0, in the call to printf(). Both versions have the same, correct behavior.
> 
> 
> <Program3>
> 
>     <foo3.c>
>     void foo3(const int* pi)
>     {
>         *(int*)pi = 1;
>     }
>     </foo3.c>
> 
>     <main3.c>
>     #include <stdio.h>
>     void foo3(const int* pi);
>     int main(void)
>     {
>         const int i = 0;
>         foo3(&i);
>         printf("i = %d\n", i);
>         return 0;
>     }
>     </main3.c>
> 
> </Program3>
> 
> Program3 defines 'i' const, and attempts to modify it through a const pointer, by casting const away in foo3().
> 
> On my particular system, when compiling Program3 with gcc 4.4.7 with optimizations turned off (-O0), the program prints "i = 1". With optimizations turned on (-O3), it prints "i = 0".
> 
> The question of which of these two behaviors is "correct" would be pointless, since Program3 has undefined behavior.
> 
> 
> Antoine
> --


-- 
Chen Gang

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