Fix rx `regexp` form with deprecated syntax

The argument of the rx `regexp` form is assumed to evaluate to a valid
regexp, but certain kinds of deprecated but still accepted usage were
not handled correctly, such as unescaped literal (special) characters:
 (rx "a" (regexp "*")) => "a*" which is wrong.
Handle these cases; there is no extra trouble.

* lisp/emacs-lisp/rx.el (rx--translate-regexp): Force bracketing
of single special characters.
* test/lisp/emacs-lisp/rx-tests.el (rx-regexp): Add test case.

2 files changed, 3 insertions, 1 deletions

diff --git a/lisp/emacs-lisp/rx.el b/lisp/emacs-lisp/rx.el
index b29b870061d..58584f300c9 100644
--- a/lisp/emacs-lisp/rx.el
+++ b/lisp/emacs-lisp/rx.el
@@ -890,7 +890,7 @@ Return (REGEXP . PRECEDENCE)."
                                (* (or (seq "[:" (+ (any "a-z")) ":]")
                                       (not (any "]"))))
                                "]")
-                          anything
+                          (not (any "*+?^$[\\"))
                           (seq "\\"
                                (or anything
                                    (seq (any "sScC_") anything)
diff --git a/test/lisp/emacs-lisp/rx-tests.el b/test/lisp/emacs-lisp/rx-tests.el
index 63d7c7b91ea..388c5e86b4c 100644
--- a/test/lisp/emacs-lisp/rx-tests.el
+++ b/test/lisp/emacs-lisp/rx-tests.el
@@ -391,6 +391,8 @@
   (let ((x "a*"))
     (should (equal (rx (regexp x) "b")
                    "\\(?:a*\\)b"))
+    (should (equal (rx "a" (regexp "*"))
+                   "a\\(?:*\\)"))
     (should (equal (rx "" (regexp x) (eval ""))
                    "a*"))))