On Mon, Aug 26, 2013 at 03:23:16PM -0700, Alex Wang wrote:
> I found answer from here:
> http://stackoverflow.com/questions/337449/how-does-one-declare-an-array-of-constant-function-pointers-in-c
>
> so if i understand it correctly:
>
> extern int (*build_assert(void))[1];
>
> declare an extern function pointer 'build_assert', pointing to a function
> that takes no input argument and returns int[1]. Right?
Not quite. You can use the "cdecl" program to see what this is:
blp@sigsegv:~/ovs/_build$ cdecl
Type `help' or `?' for help
cdecl> explain extern int (*build_assert(void))[1];
declare build_assert as extern function (void) returning pointer to array 1
of int
cdecl>
blp@sigsegv:~/ovs/_build$
> But I'm still not clear why we must use (*build_assert(void))? I tried to
> change it to normal variable declaration like below:
> /* Build-time assertion for use in a declaration context. */
> #define BUILD_ASSERT_DECL(EXPR) \
> /* Build-time assertion for use in a declaration context. */
> #define BUILD_ASSERT_DECL(EXPR) \
> - extern int (*build_assert(void))[BUILD_ASSERT__(EXPR)]
> + extern int build_assert[BUILD_ASSERT__(EXPR)]
>
> But I got "unused variable" warnings. So, is using "(*build_assert(void))"
> for suppressing the "unused variable" warnings only?
Yes, compilers do not ordinarily warn about functions that are
declared but not called.
The ideas behind BUILD_ASSERT come from gnulib. You can read the
extensive documentation here:
http://git.savannah.gnu.org/gitweb/?p=gnulib.git;a=blob;f=lib/verify.h;h=d42d0750ee13f2b1db1274b7db14079b77e8dbef;hb=HEAD
The history of that file is also informative.
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