"6.6 Constant expressions" doesn't allow a cast from a non-arithmetic type
to an arithmetic one generally, and a cast
in an address constant can only cast from an integer constant to a pointer
type (eg, char *reg = (char*)0x123450);

the one example with 8c escaped with a warning ("initialize pointer to an
integer") because of some 8c x86-specific folding that makes the expression
acceptable.

even so, the format and intention of the example seems practical (with the
correct cast to uintptr) and "An implementation may accept other forms of
constant expressions".
it should be fairly easy to add as an extension with consistent handling
across ?c.



On Wed, 1 Aug 2018 at 08:22, Bakul Shah <bakul@bitblocks.com> wrote:

> Consider:
>
> % cat x.c
> #include <u.h>
> uintptr foo[3];
> uintptr bar=&foo[2];
>
> % 8c -c x.c     # this works.
> % 5c -c x.c     # this fails
> x.c:3 initializer is not a constant: bar
>
> If I change the last line to
>
> uintptr* bar=&foo[2];
>
> Both compilers compile it fine. But if I change the last line
> to
>
> uintptr bar=(uintptr)&foo[2];
>
> both compilers fail. Note that the last two examples are
> type correct.
>
> uintptr is the same size as int* so there should be no runtime
> cost and we already know from the second example that &foo[2]
> is a link time constant.
>
> Similar code to the last two examples compiles on gcc/clang.
>
> This seems like a compiler bug.
>
>