It looks like we have more work to do and they are not ready. Thanks
________________________________
From: Uwe Brauer
Sent: Tuesday, May 21, 2024 9:39 AM
To: John Ciolfi
Cc: Eric Ludlam; Uwe Brauer; Matlab-emacs-discuss
Subject: Re: [Matlab-emacs-discuss] the filling patch and its problems
"JC" == John Ciolfi <cio...@mathworks.com> writes:
Hi John
Hi Uwe, I don't think these fill-paragraph patches made it into master.
Attached are the patches that work for me. Are there any reasons to not apply
them?
In July last year Eric provided some patches for improving the filling.
I started a new branch and run some test and sooner or later them there were
corner cases where it did not work as expected or made the filling worse.
Your patch differ from all the commits, so I just applied it to master and
start testing. However there also cases where there problems.
The following is taken from files from Students who wrote their code using the
native matlab editor in MS Windows. I give you some examples, now. But if you
wish I push your patch as a different branch, and add a directory
tests-for-new-filling or so.
Number 1
j_func = @(t, y_1, jfunc, h, y_2) [h*jfunc(t, y_2) - jfunc(t, y_2)\jfunc(t,
y_2)]; % UB:03.05.2024:18:08: no entiendo esta linea
% explicalo por favor lo utilizo para calcular la derivada de @func;
debería poder usar 1 en lugar de la segunda parte de la ecuacion pero me da
error y no logro ver por qué, como así me funcionaba no lo he tocado
When running matlab-fill-comment-line cursor is at the end of the first line I
obtain
j_func = @(t, y_1, jfunc, h, y_2) [h*jfunc(t, y_2) - jfunc(t, y_2)\jfunc(t,
y_2)]; %
% UB:03.05.2024:18:08:
% no entiendo
% esta linea
% explicalo por favor lo utilizo para calcular la derivada de @func;
debería poder usar 1 en lugar de la segunda parte de la ecuacion pero me da
error y no logro ver por qué, como así me funcionaba no lo he tocado
When applying the function after the word «tocado», I obtain
j_func = @(t, y_1, jfunc, h, y_2) [h*jfunc(t, y_2) - jfunc(t, y_2)\jfunc(t,
y_2)]; %
% UB:03.05.2024:18:08:
% no entiendo
% esta linea
% explicalo
% por favor
% lo utilizo
% para
% calcular la
% derivada de
% @func;
% debería
% poder usar
% 1 en lugar
% de la
% segunda
% parte de la
% ecuacion
% pero me da
% error y no
% logro ver
% por qué,
% como así me
% funcionaba
% no lo he
% tocado
And this looks wired to me, don't you think?
Uwe
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