On Mon, Jul 22, 2019 at 5:13 PM Qian Cai <cai@xxxxxx> wrote:
>
> On Fri, 2019-07-19 at 17:47 -0400, Qian Cai wrote:
> > On Thu, 2019-07-18 at 16:29 -0700, David Miller wrote:
> > > From: Qian Cai <cai@xxxxxx>
> > > Date: Thu, 18 Jul 2019 19:26:47 -0400
> > >
> > > >
> > > >
> > > > > On Jul 18, 2019, at 5:21 PM, Bill Wendling <morbo@xxxxxxxxxx> wrote:
> > > > >
> > > > > [My previous response was marked as spam...]
> > > > >
> > > > > Top-of-tree clang says that it's const:
> > > > >
> > > > > $ gcc a.c -O2 && ./a.out
> > > > > a is a const.
> > > > >
> > > > > $ clang a.c -O2 && ./a.out
> > > > > a is a const.
> > > >
> > > >
> > > >
> > > > I used clang-7.0.1. So, this is getting worse where both GCC and clang
> > > > will
> > >
> > > start to suffer the
> > > > same problem.
> > >
> > > Then rewrite the module parameter macros such that the non-constness
> > > is evident to all compilers regardless of version.
> > >
> > > That is the place to fix this, otherwise we will just be adding hacks
> > > all over the place rather than in just one spot.
> >
> > The problem is that when the compiler is compiling be_main.o, it has no
> > knowledge about what is going to happen in load_module(). The compiler can
> > only
> > see that a "const struct kernel_param_ops" "__param_ops_rx_frag_size" at the
> > time with
> >
> > __param_ops_rx_frag_size.arg = &rx_frag_size
> >
> > but only in load_module()->parse_args()->parse_one()->param_set_ushort(), it
> > changes "__param_ops_rx_frag_size.arg" which in-turn changes the value
> > of "rx_frag_size".
>
> Even for an obvious case, the compilers still go ahead optimizing a variable as
> a constant. Maybe it is best to revert the commit d66acc39c7ce ("bitops:
> Optimise get_order()") unless some compiler experts could improve the situation.
>
> #include <stdio.h>
>
> int a = 1;
>
> int main(void)
> {
> int *p;
>
> p = &a;
> *p = 2;
>
> if (__builtin_constant_p(a))
> printf("a is a const.\n");
>
> printf("a = %d\n", a);
>
> return 0;
> }
>
> # gcc -O2 const.c -o const
>
> # ./const
> a is a const.
> a = 2
This example (like the former) is showing correct behavior. At the
point of invocation of __builtin_constant_p here, the compiler knows
that 'a' is 2, because you've just assigned it (through 'p', but that
indirection trivially disappears in optimization).
