>>> On 10.08.16 at 11:46, <jbeul...@suse.com> wrote:
> Odd. I've tried this simple example:
>
> typedef int fn_t(void);
>
> struct s {
> unsigned n;
> fn_t**fn;
> fn_t*const*fnc;
> const fn_t**cfn;
> };
>
> int test1(const struct s*ps) {
> unsigned i;
> int rc = 0;
>
> for(i = 0; !rc && i < ps->n; ++i)
> rc = ps->fn[i]();
>
> return rc;
> }
>
> int test2(const struct s*ps) {
> unsigned i;
> int rc = 0;
>
> for(i = 0; !rc && i < ps->n; ++i)
> rc = ps->fnc[i]();
>
> return rc;
> }
>
> int test3(const struct s*ps) {
> unsigned i;
> int rc = 0;
>
> for(i = 0; !rc && i < ps->n; ++i)
> rc = ps->cfn[i]();
>
> return rc;
> }
>
> test1() and test2() get compiled identically. test3(), using the field
> with the misplaced const, oddly enough gets compiled slightly
> differently (and without a warning despite one would seem
> warranted), yet the call doesn't get omitted. If, however, I change
> the return type of fn_t to void, the function body of test3() ends
> up empty, which is a compiler bug afaict, but which also suggests
> that you've tried the variant with the misplaced const.
FTR: This is not a compiler bug, as specifically named undefined
in the C spec.
Jan
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