On 16/10/2022 23:33, Juan Manuel Macías wrote:
(replace-regexp-in-string
"^[ \t]+" (lambda (m) (format "\\hspace*{%dem}" (length m)))
(replace-regexp-in-string
"\\(\\\\\\\\\n\\)+\\([ \t]*\\\\\\\\\\)+" "\n"
(replace-regexp-in-string
"\\([ \t]*\\\\\\\\\\)?[ \t]*\n" "\\\\\n"
(setq contents
(if lin
(replace-regexp-in-string "\\(\n\\)+\\([ \t]*\n\\)+"
"\\\\\\\\!\n\n"
contents)
contents)) nil t) nil t) nil t) linreset)
I had a hope, it is possible to do it in a single pass of
`replace-regexp-in-string', but unfortunately the function does not
allow to make conditional substitution based on (rx (optional (group
string-start))) (a bug?).
I still prefer to avoid replacement of latex newlines back to empty
string. Though I am really happy with the following code, I expected a
more concise snippet. Unit tests may found bugs in it.
(let ((contents "\n\n 0 \n\n\na b\nc d e \n\n\nf g \n h i\n\n"))
;; Strip leading newlines.
(setq contents
(replace-regexp-in-string
(rx string-start (1+ (0+ blank) ?\n)) ""
contents 'fixed-case 'literal))
;; Add explicit line breaks and strip trailing spaces.
(setq contents
(replace-regexp-in-string
(rx (0+ blank) ?\n
(optional (group (1+ (0+ blank) ?\n)))
(optional (group (0+ blank) (not (any blank ?\n)))))
(lambda (m)
(let ((par (match-string 1 m))
(next (match-string 2 m)))
(if next
(concat (if par "\n\n" "\\\\\n")
next)
"")))
contents 'fixed-case 'literal))
;; Indented lines.
(replace-regexp-in-string
(rx line-start (1+ blank))
(lambda (m) (format "\\hspace*{%dem}" (length m)))
contents 'fixed-case 'literal))
Feel free to use it for inspiration during your work on a patch.
