Escaping the `repl` argument to `re.sub()`, `re.subn()`

Documentation

It's not immediately obvious how to escape the repl (replacement for matches) argument to re.sub() and re.subn() if repl is chosen by a potentially hostile actor. Obviously, re.escape() isn't the answer, as that escapes far too much.

The right answer seems to be escaped_repl = raw_repl.replace(bslash, bslash*2) where bslash = '\\'. It might be worth adding this to the documentation.

Here's the code I used to empirically validate the "right answer" given above (checked on Python 3.8 & 3.12):

from __future__ import annotations
import re, sys

def escape_re_sub_repl(repl: str) -> str:

    return repl.replace('\\', '\\\\')

def test_escape_re_sub_repl() -> None:

    backslash = '\\'
    assert len(backslash) == 1

    base_regex = 'TARGET'
    assert base_regex == re.escape(base_regex)
    base_prefix = 'BEFORE:'
    base_suffix = ':AFTER'
    base_input = f'{base_prefix}{base_regex}{base_suffix}'

    base_chars = tuple(chr(p) for p in range(sys.maxunicode + 1))
    escaped_chars = tuple(f'{backslash}{c}' for c in base_chars)
    test_cases = base_chars + escaped_chars
    assert {len(f) for f in test_cases} == {1, 2}

    for raw in test_cases:
        repl = escape_re_sub_repl(raw)
        got, change_count = re.subn(base_regex, repl, base_input)
        assert change_count == 1
        assert got == f'{base_prefix}{raw}{base_suffix}'

Beginne damit, die Dokumentation zu re.sub() und re.subn() sowie deren Regeln für Ersetzungszeichenfolgen zu finden. Prüfe die vorhandenen Hinweise zu re.escape() und dokumentiere anschließend den Ansatz zur Verdopplung von Backslashes für problematischen Ersetzungstext; validiere ihn anhand der bereitgestellten Beispiele. Als abgeschlossen gilt die Aufgabe, wenn die Hinweise zum sicheren Escaping in der relevanten Dokumentation klar sind.