Re: [PATCH v7 7/8] fortify: Make sure strlen() may still be used as a constant expression

From: Kees Cook
Date: Tue Feb 08 2022 - 18:58:32 EST


On Tue, Feb 08, 2022 at 03:17:13PM -0800, Nick Desaulniers wrote:
> On Tue, Feb 8, 2022 at 2:53 PM Kees Cook <keescook@xxxxxxxxxxxx> wrote:
> >
> > In preparation for enabling Clang FORTIFY_SOURCE support, redefine
> > strlen() as a macro that tests for being a constant expression
> > so that strlen() can still be used in static initializers, which is
> > lost when adding __pass_object_size and __overloadable.
> >
> > An example of this usage can be seen here:
> > https://lore.kernel.org/all/202201252321.dRmWZ8wW-lkp@xxxxxxxxx/
> >
> > Notably, this constant expression feature of strlen() is not available
> > for architectures that build with -ffreestanding. This means the kernel
> > currently does not universally expect strlen() to be used this way, but
> > since there _are_ some build configurations that depend on it, retain
> > the characteristic for Clang FORTIFY_SOURCE builds too.
> >
> > Signed-off-by: Kees Cook <keescook@xxxxxxxxxxxx>
> > ---
> > include/linux/fortify-string.h | 13 +++++++++++--
> > 1 file changed, 11 insertions(+), 2 deletions(-)
> >
> > diff --git a/include/linux/fortify-string.h b/include/linux/fortify-string.h
> > index db1ad1c1c79a..f77cf22e2d60 100644
> > --- a/include/linux/fortify-string.h
> > +++ b/include/linux/fortify-string.h
> > @@ -2,6 +2,8 @@
> > #ifndef _LINUX_FORTIFY_STRING_H_
> > #define _LINUX_FORTIFY_STRING_H_
> >
> > +#include <linux/const.h>
> > +
> > #define __FORTIFY_INLINE extern __always_inline __gnu_inline
> > #define __RENAME(x) __asm__(#x)
> >
> > @@ -95,9 +97,16 @@ __FORTIFY_INLINE __kernel_size_t strnlen(const char * const p, __kernel_size_t m
> > return ret;
> > }
> >
> > -/* defined after fortified strnlen to reuse it. */
> > +/*
> > + * Defined after fortified strnlen to reuse it. However, it must still be
> > + * possible for strlen() to be used on compile-time strings for use in
> > + * static initializers (i.e. as a constant expression).
> > + */
> > +#define strlen(p) \
> > + __builtin_choose_expr(__is_constexpr(__builtin_strlen(p)), \
>
> Is `__is_constexpr(p) == __is_constexpr(__builtin_strlen(p))`? i.e.
> can we drop the first `__builtin_strlen`? It seems redundant.
>
> So instead, we'd have:
>
> #define strlen(p) __builtin_choose_expr(__is_constexpr(p),
> __builtin_strlen(p), __fortify_strlen(p))
>
> Or is there some funny business where p isn't constexpr but strlen(p)
> somehow is? I doubt that. (Or is it that p is constexpr, but
> strlen(p) is not?)
>
> (Guess I'm wrong: https://godbolt.org/z/19ffz7vjx)

Yeah, as you've discovered ... funny business. :P

> Ok then.
> Reviewed-by: Nick Desaulniers <ndesaulniers@xxxxxxxxxx>

Thanks!

--
Kees Cook