Messages in this thread | | | Date | Sun, 13 Jan 2013 16:10:08 +0800 | From | Chen Gang F T <> | Subject | Re: Why is the kfree() argument const? |
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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 *).
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
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