(buffer-substring-no-properties) backslash handling

Question

Good afternoon. In the text, the record

@samp{\}

, the code

(buffer-substring-no-properties (+ nach 4) (- (point) 4))

is captured here this region, at the output of the

"@samp{\\}"

function. The question is how to get the non-escaped forward slash. Thank you

Answer 1

Changes further along the code gave the result, namely, not a correction of the record of the read line, but a change in the insertion of the ill-fated line "@samp {\\}" back into the text. The function is presented

(defun zamenaOBJ (спсОбктв)
  "Зменить obzx оригиналом"
  (goto-char 1)  
  (while (re-search-forward "[Bb]jkt\\([[:digit:]]\\)+" nil t 1)
    (replace-match (cdr (assoc (downcase (match-string 0)) спсОбктв)) t)))

Here is the data at the time of the error (what was received

(buffer-substring-no-properties)

and saved in the associative list

(cdr (assoc (downcase (match-string 0)) спсОбктв))
"@samp{\\}"

The error itself

(replace-match (cdr (assoc (downcase (match-string 0)) спсОбктв)) t)

Invalid use of ‘\’ in replacement text

If it was written like this:

(replace-match (prin1-to-string (cdr (assoc (downcase (match-string 0)) спсОбктв)) t))

все замены по тексту были окружены двойной кавычкой, если же указывался аргумент noescape в t для prin1-to-string:

(replace-match (prin1-to-string (cdr (assoc (downcase (match-string 0)) спсОбктв)) t) t)

or

(replace-match (princ (cdr (assoc (downcase (match-string 0)) спсОбктв)) t))

the error came back again

Solved by adding the second argument literal replace-match

(replace-match (cdr (assoc (downcase (match-string 0)) спсОбктв)) t t)