Re: [PATCH v4 1/2] lib/test_bitops: Add benchmark test for fns()

[Yury Norov <yury.norov@gmail.com>]

On Wed, May 01, 2024 at 09:20:46PM +0800, Kuan-Wei Chiu wrote:
> Introduce a benchmark test for the fns(). It measures the total time
> taken by fns() to process 1,000,000 test data generated using
> get_random_bytes() for each n in the range [0, BITS_PER_LONG).
> 
> example:
> test_bitops: fns:          5876762553 ns, 64000000 iterations

So... 5 seconds for a test sounds too much. I see the following patch
improves it dramatically, but in general let's stay in a range of
milliseconds. On other machines it may run much slower and trigger
a stall watchdog.

> Signed-off-by: Kuan-Wei Chiu <visitorckw@gmail.com>

Suggested-by: Yury Norov <yury.norov@gmail.com>

> ---
> 
> Changes in v4:
> - Correct get_random_long() -> get_random_bytes() in the commit
>   message.
> 
>  lib/test_bitops.c | 22 ++++++++++++++++++++++
>  1 file changed, 22 insertions(+)
> 
> diff --git a/lib/test_bitops.c b/lib/test_bitops.c
> index 3b7bcbee84db..ed939f124417 100644
> --- a/lib/test_bitops.c
> +++ b/lib/test_bitops.c
> @@ -50,6 +50,26 @@ static unsigned long order_comb_long[][2] = {
>  };
>  #endif
>  
> +static unsigned long buf[1000000];

Can you make it __init, or allocate with kmalloc_array(), so that 64M
of memory will not last forever in the kernel?

> +static int __init test_fns(void)
> +{
> +	unsigned int i, n;
> +	ktime_t time;
> +
> +	get_random_bytes(buf, sizeof(buf));
> +	time = ktime_get();
> +
> +	for (n = 0; n < BITS_PER_LONG; n++)
> +		for (i = 0; i < 1000000; i++)
> +			fns(buf[i], n);

What concerns me here is that fns() is a in fact a const function, and
the whole loop may be eliminated. Can you make sure it's not your case
because 450x performance boost sounds a bit too much to me.

You can declare a "static volatile __used __init" variable to assign
the result of fns(), and ensure that the code is not eliminated

> +	time = ktime_get() - time;
> +	pr_err("fns:  %18llu ns, %6d iterations\n", time, BITS_PER_LONG * 1000000);

Here the number of iterations is always the same. What's the point to
print it?

Thanks,
Yury