bug#60691: 29.0.60; Slow tree-sitter font-lock in ruby-ts-mode

[Dmitry Gutov]

Hi Yuan,

On 29/01/2023 10:25, Yuan Fu wrote:

> > > > So if previously it happened once somehow during a certain scenario, now I have 
> > > > to repeat the same scenario 4 times, and the condition is met.
> > > I was hoping that the scenario only happen once, oh well :-) I’ll
> > > change the decision based on analyzing the tree’s dimension: too
> > > deep or too wide activates the fast mode. Let’s see how it works.
> >
> > Thank you, let me know when it's time to test again.
>
> Sorry for the delay. Now treesit-font-lock-fontify-region uses
> treesit-subtree-stat to determine whether to enable the "fast mode". Now
> it should be impossible to activate the fast mode on moderately sized
> buffers.

Thank you, it seems to work just fine in my scenario. And 
treesit-subtree-stat makes sense.

I have a few more questions about the current strategy, though.

IIUC, we only do the treesit--font-lock-fast-mode test once in 
treesit-font-lock-fontify-region, and then use the detected value for 
the whole later life of the buffer. Is that right?

What if the buffer didn't originally have the problematic error nodes we 
are guarding from, and then later the user wrote enough code to have at 
least one of them? If they didn't close Emacs, or revert the buffer, our 
logic still wouldn't use the "fast node", would it?

Or vice versa: if the buffer started out with error nodes, and 
consequently, "fast mode", but then the user has edited it so that those 
error nodes disappeared, shouldn't the buffer stop using the "fast mode"?

From my measurements, in ruby-mode, at least treesit-subtree-stat is 
20-40x faster than refontifying the whole buffer. So one possible 
strategy would be to repeat the test every time. I'm not sure it's fast 
enough in the "problem" buffers, though, and I don't have any to test.

In those I did test, though, it takes ~1 ms.

But we could repeat the test only once every couple of seconds and/or 
after the buffer has changed again. That would hopefully make it a 
non-bottleneck in all cases.