Re: [PATCH v4 2/3] prandom: Add __always_inline version of prandom_u32_state()

["Jason A. Donenfeld" <Jason@zx2c4.com>]

On Mon, Jan 19, 2026 at 01:01:09PM +0000, Ryan Roberts wrote:
> We will shortly use prandom_u32_state() to implement kstack offset
> randomization and some arches need to call it from non-instrumentable
> context. So let's implement prandom_u32_state() as an out-of-line
> wrapper around a new __always_inline prandom_u32_state_inline(). kstack
> offset randomization will use this new version.
> 
> Acked-by: Mark Rutland <mark.rutland@arm.com>
> Signed-off-by: Ryan Roberts <ryan.roberts@arm.com>
> ---
>  include/linux/prandom.h | 20 ++++++++++++++++++++
>  lib/random32.c          |  8 +-------
>  2 files changed, 21 insertions(+), 7 deletions(-)
> 
> diff --git a/include/linux/prandom.h b/include/linux/prandom.h
> index ff7dcc3fa105..801188680a29 100644
> --- a/include/linux/prandom.h
> +++ b/include/linux/prandom.h
> @@ -17,6 +17,26 @@ struct rnd_state {
>  	__u32 s1, s2, s3, s4;
>  };
>  
> +/**
> + * prandom_u32_state_inline - seeded pseudo-random number generator.
> + * @state: pointer to state structure holding seeded state.
> + *
> + * This is used for pseudo-randomness with no outside seeding.
> + * For more random results, use get_random_u32().
> + * For use only where the out-of-line version, prandom_u32_state(), cannot be
> + * used (e.g. noinstr code).
> + */
> +static __always_inline u32 prandom_u32_state_inline(struct rnd_state *state)

This is pretty bikesheddy and I'm not really entirely convinced that my
intuition is correct here, but I thought I should at least ask. Do you
think this would be better called __prandom_u32_state(), where the "__"
is kind of a, "don't use this directly unless you know what you're doing
because it's sort of internal"? It seems like either we make this inline
for everybody, or if there's a good reason for having most users use the
non-inline version, then we should be careful that new users don't use
the inline version. I was thinking the __ would help with that.

Jason