$world = "WΩrlδ"
"(Hell)*o $world\n"           # <--- plain (Hell)*o,resolve $world,escape \n
'(Hell)*o $world\n'           # <--- plain (Hell)*o,plain $world,escape \n
/(Hell)*o $world\n/           # <--- regexp (Hell)*,interpret \n
<(Hell)*o $world\n>           # <--- make list ["(Hello*o","$world\n"]
{(Hell)*o $world\n}           # <--- Syntax error,this language cant' parse it

Perl 6也足够强大,可以在未来的语言中存在类似的东西

my $emacs_func = (defun perl-backward-to-start-of-continued-exp (lim)
      (if (= (preceding-char) ?\))
          (forward-sexp -1))
      (beginning-of-line)
      (if (<= (point) lim)
          (goto-char (1+ lim)))
      (skip-chars-forward " \t\f"))


$ typeof($emacs_func)
> Emacs Lisp list

所以,问题显然是：它可以在Perl 6的当前规范(甚至实现)中完成吗？