I tried this -
---------------------------------
struct list_head {
struct list_head *next, *prev;
};
#define INIT_LIST_HEAD(ptr) {(ptr)->next = (ptr); (ptr)->prev = (ptr);}
int main ()
{
int i;
struct list_head mylist;
struct list_head *ptr = &mylist;
printf ("hi\n");
if (i)
{
INIT_LIST_HEAD(ptr);
}
else
{
printf ("jkldkas\n");
}
return 0;
}
--------------------------
And there is no compilation error.
On 12/26/07, Li Zefan <[EMAIL PROTECTED]> wrote:
>
> sahlot arvind wrote:
> > Recently I started looking into linux kernel and trying to understand
> the
> > code.
> > I am working with linux-2.6.9.
> > in file include/llinux/list.h - I found something like this.
> >
> > #define INIT_LIST_HEAD(ptr) do { \
> > (ptr)->next = (ptr); (ptr)->prev = (ptr); \
> > } while (0)
> >
> >
> > My question is why do we use a loop when we actually know that it is not
> > going to execute more than once? Cannot we simply do -
> >
> > #define INIT_LIST_HEAD(ptr) {(ptr)->next = (ptr); (ptr)->prev =
> (ptr)}
> >
> > Do we get some kind of optimization by using while (0)?
> >
>
> with #define INIT_LIST_HEAD(ptr) {(ptr)->next = (ptr); (ptr)->prev =
> (ptr);}
>
> try this:
>
> if (...)
> INIT_LIST_HEAD(ptr);
> else
> ...;
>
> you will get compilation error:
>
> error: expected expression before 'else'
>
> see why? expand the macro:
>
> if (...)
> {
> (ptr)->next = (ptr);
> (ptr)->prev = (ptr);
> }; <---here
> else
> ...;
>
