Messages in this thread | | | Date | Sat, 12 Jan 2013 11:18:37 -0800 (PST) | Subject | Re: Why is the kfree() argument const? | From | antoine.trux@gmail ... |
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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
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