Re: [RFC PATCH] preempt: Add a generic function to return the preemption string.

[Steven Rostedt]

On Thu, 9 Jan 2025 15:37:01 +0100
Peter Zijlstra <peterz@xxxxxxxxxxxxx> wrote:

> > I'm sorry, but I can't even tell what the above is doing without my brain
> > hurting. Why make code that was easy to read into a cryptic obfuscation? I
> > can't see this as an optimization as IS_ENABLED() is determined at compile
> > time.  
> 
> Upgrade brain. Also the proposed thing was just plain wrong.

Or use seq_buf to make it a little more readable again:

const char *preempt_model_str(void)
{
	static char buf[128];
	bool brace = IS_ENABLED(CONFIG_PREEMPT_RT) &&
		     (IS_ENABLED(CONFIG_PREEMPT_DYNAMIC) ||
		      IS_ENABLED(CONFIG_PREEMPT_LAZY));

	if (IS_ENABLED(CONFIG_PREEMPT_BUILD)) {
		struct seq_buf s;

		seq_buf_init(&s, buf, 128);
		seq_buf_puts(&s, "PREEMPT");

		if (IS_ENABLED(CONFIG_PREEMPT_RT))
			seq_buf_printf(&s, "%sRT%s", brace ? "_{" : "_",
					brace ? "," : "");

		if (IS_ENABLED(CONFIG_PREEMPT_DYNAMIC)) {
			seq_buf_printf(&s, "(%s)%s",
				preempt_modes[preempt_dynamic_mode],
				brace ? "}" : "");
			return seq_buf_str(&s);
		}

		if (IS_ENABLED(CONFIG_PREEMPT_LAZY)) {
			seq_buf_printf(&s, "LAZY%s",
				brace ? "}" : "");
			return seq_buf_str(&s);
		}

		return seq_buf_str(&s);
	}

	if (IS_ENABLED(CONFIG_VOLUNTARY_BUILD))
		return "VOLUNTARY";

	return "NONE";
}


At least this removes the need for keeping track of off, len and r.

-- Steve