Re: [PATCH v1 2/2] rust: Add read_poll_timeout_atomic function

["Danilo Krummrich" <dakr@kernel.org>]

On Thu Aug 21, 2025 at 5:57 AM CEST, FUJITA Tomonori wrote:
> +pub fn read_poll_timeout_atomic<Op, Cond, T>(
> +    mut op: Op,
> +    mut cond: Cond,
> +    delay_delta: Delta,
> +    timeout_delta: Delta,
> +) -> Result<T>
> +where
> +    Op: FnMut() -> Result<T>,
> +    Cond: FnMut(&T) -> bool,
> +{
> +    let mut left_ns = timeout_delta.as_nanos();
> +    let delay_ns = delay_delta.as_nanos();
> +
> +    loop {
> +        let val = op()?;
> +        if cond(&val) {
> +            // Unlike the C version, we immediately return.
> +            // We know the condition is met so we don't need to check again.
> +            return Ok(val);
> +        }
> +
> +        if left_ns < 0 {
> +            // Unlike the C version, we immediately return.
> +            // We have just called `op()` so we don't need to call it again.
> +            return Err(ETIMEDOUT);
> +        }
> +
> +        if !delay_delta.is_zero() {
> +            udelay(delay_delta);
> +            left_ns -= delay_ns;
> +        }
> +
> +        cpu_relax();
> +        left_ns -= 1;

How do we know that each iteration costs 1ns? To make it even more obvious, we
don't control the implementation of cond(). Shouldn't we use ktime for this?