odoc-parser: strip stray CR from deindented code block lines

Splitting a code block on '\n' leaves a trailing '\r' on every line of
CRLF input. count_leading_whitespace already discounts that '\r' when
computing the indent, but drop didn't strip it from the line it returned,
so the joined result still contained CRs.

This made fmt_code on Windows fail with Lexer.Error (Illegal_character
'\r') for any {@ocaml[...]} block, since the OCaml lexer's newline regex
is '\r*\n' and rejects a bare CR.

Strip the trailing '\r' in drop so the output uses LF only, matching
what the rest of the pipeline expects.

Author: hhugo

CHANGES.md

@@ -14,6 +14,9 @@ profile. This started with version 0.26.0.
 - Fix `match` in `if-then-else` branches expanding to vertical with
   `fit-or-vertical` (#2798, @MisterDA)
 
+- Fix lexer error on Windows when an Odoc code block contains CRLF line
+  endings (@hhugo)
+
 ## 0.29.0
 
 ### Highlight

vendor/odoc-parser/odoc_parser.ml

@@ -168,11 +168,13 @@ let deindent : what:string -> loc:Loc.span -> string -> string * Warning.t list
     else []
   in
   let drop n line =
-    (* Since blank lines were ignored when calculating
-       [least_amount_of_whitespace], their length might be less than the
-       amount. *)
-    if String.length line < n then ""
-    else String.sub line n (String.length line - n)
+    (* Strip a trailing '\r' left over from splitting CRLF input on '\n', so
+       the joined output uses LF only and downstream consumers (e.g. the OCaml
+       lexer, whose newline regex is '\r*\n') don't see a bare '\r'. Blank
+       lines may also be shorter than [n], in which case the result is empty. *)
+    let len = String.length line in
+    let stop = if len > 0 && line.[len - 1] = '\r' then len - 1 else len in
+    if n >= stop then "" else String.sub line n (stop - n)
   in
   let lines = List.map (drop least_amount_of_whitespace) lines in
   (String.concat "\n" lines, warning)