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SubjectRe: Why is the kfree() argument const?
From
On Sun, Jan 13, 2013 at 9:10 AM, Chen Gang F T
<chen.gang.flying.transformer@gmail.com> wrote:
> 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 *).
you are definitely wrong. both of them are correct - it's the
difference between kernel space and user space.

>
> 于 2013年01月13日 03:18, antoine.trux@gmail.com 写道:
>> 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
>
> Flying Transformer
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