Re: How to get a concatenation of the negations with rx (ex: [^a][^b])?

[Michael Heerdegen <michael_heerdegen@web.de>]

Edgar Lux <edgarlux@mailfence.com> writes:

> Hello. I am trying to get this regular expression:
>
>     "[^a][^b]"
>
> in an easier way. I thought that I could do 
>
>     (rx (not (seq "a" "b")))

No,

> but that got me 
>
>     Debugger entered--Lisp error: (error "Illegal argument to rx
> ‘not’: (seq \"a\" \"b\")")

this would also express something different: `not' is not distributive
over `seq' concatenation of regexps -- a string starting with a character
different from 'a' followed by a character different from 'b' is something
different than a string not starting with "ab".

> The error is very clear,

I think `match a string not starting with "ab"' is not even expressible
using a standalone standard regexp.  That would require regexps to be
able to backtrack when matching, and it would be something more general
than regexps.

> but I would like to know if there is a smart way of achieving the same
> without having to type:
>
>     (rx (seq (not "a") (not "b")))

Unless you actually wanted to express something different: the `seq'
wrapper is redundant, but with

  (rx (not "a") (not "b"))

a local maximum of smartness is reached.


Michael.