Re: [PATCH v3] random: handle archrandom with multiple longs

[Jason A. Donenfeld]

Hi Boris,

On Mon, Jul 25, 2022 at 11:19:01AM +0200, Borislav Petkov wrote:
> On Tue, Jul 19, 2022 at 03:02:07PM +0200, Jason A. Donenfeld wrote:
> > Since callers need to check this return value and loop anyway, each arch
> > implementation does not bother implementing its own loop to try again to
> > fill the maximum number of longs. Additionally, all existing callers
> > pass in a constant max_longs parameter.
> 
> Hmm, maybe this has come up already but it reads weird.
> 
> If I have a function arch_get_random_longs(), I'd expect it to give me
> the number of longs I requested or say, error.
> 
> Why do the callers need to loop?
> 
> If I have to loop, I'd call the "get me one long" function and loop N
> times.

Answered partially in the commit message you quoted and partially here:
https://lore.kernel.org/lkml/YtqIbrds53EuyqPE@xxxxxxxxx/

Note that arch_get_random_longs() is not a general purpose function. For
that there used to be get_random_bytes_arch(), but that no longer exists
as people shouldn't be using this stuff directly. arch_get_random_longs()
is a special purpose function mainly intended for use by the RNG itself.

More directly, the reason we don't want to error is because the use case
has fallbacks meant to handle errors. The cascade looks like this
(quoting from the other email):

    unsigned long array[whatever];
    for (i = 0; i < ARRAY_SIZE(array);) {
        longs = arch_get_random_seed_longs(&array[i], ARRAY_SIZE(array) - i);
        if (longs) {
            i += longs;
            continue;
        }
        longs = arch_get_random_longs(&array[i], ARRAY_SIZE(array) - i);
        if (longs) {
            i += longs;
            continue;
        }
        array[i++] = random_get_entropy();
    }

It tries to get the best that it can as much as it can, but isn't going
to block or do anything too nuts for that.

Anyway, from an x86 perspective, I can't imagine you object to this
change, right? Codegen is the same.

Jason