On Tue, 10 Mar 2015, Robbert Krebbers wrote:
> On 03/10/2015 05:44 PM, Robbert Krebbers wrote:
> > On 03/10/2015 05:18 PM, Martin Sebor wrote:
> > > I suspect every compiler relies on this requirement in certain
> > > cases otherwise copying would require making use of temporary
> > > storage. Here's an example:
> > Thanks, this example is indeed not already undefined by effective types,
> > nor 6.2.6.1p6.
> Now to make it more subtle. As far as I understand 6.5.16.1p3, undefined
> behavior can already occur without the use of union types or malloc. For
> example:
>
> struct S { int x, y; };
> int main() {
> struct S p = (struct S){ .x = 10, .y = 12 };
> p = (struct S){ .x = p.x, .y = 13 };
> return p.x;
> }
>
> is undefined AFAIK.
I don't see why that would be undefined. The value stored in p isn't read
from something overlapping with p, it's read from the compound literal
object that in turn was initialized with { .x = p.x, .y = 13 }.
Similarly, in the union case in
<https://gcc.gnu.org/ml/gcc/2015-03/msg00092.html>, if you had
struct A f (struct A arg) { return arg; }
and did "u.b.b2 = f (u.a);" instead of "u.b.b2 = u.a;", that would not be
undefined (see 6.8.6.4 and GCC PR 43784).
--
Joseph S. Myers
jos...@codesourcery.com
