Should not * be greedy in a regexp?

Should not * be greedy in a regexp?

[Lennart Borgman (gmail)]

Starting from

    emacs -Q

Put this in the *scratch* buffer and eval it:

(let* ((mod-regx+ "\\(\\(?:[CSM]-\\)+\\)")
        (mod-regx* "\\(\\(?:[CSM]-\\)*\\)")
        (str "<C-S-tab>")
        (m+ (when (string-match mod-regx+ str)
              (match-string 0 str)))
        (m* (when (string-match mod-regx* str)
              (match-string 0 str))))
   (lwarn 't :warning "m+=%s, m*=%s" m+ m*))

I get

   Warning (t): m+=C-S-, m*=

Should not both this regexp give the same result here? It looks to me 
like * is treated as *? - ie non-greedy.


In GNU Emacs 22.1.50.1 (i386-mingw-nt5.1.2600)
  of 2007-07-04
Windowing system distributor `Microsoft Corp.', version 5.1.2600
configured using `configure --with-gcc (3.4) --cflags -Ic:/g/include'

Re: Should not * be greedy in a regexp?

[Andreas Schwab]

"Lennart Borgman (gmail)" <lennart.borgman@gmail.com> writes:

> Starting from
>
>    emacs -Q
>
> Put this in the *scratch* buffer and eval it:
>
> (let* ((mod-regx+ "\\(\\(?:[CSM]-\\)+\\)")
>        (mod-regx* "\\(\\(?:[CSM]-\\)*\\)")
>        (str "<C-S-tab>")
>        (m+ (when (string-match mod-regx+ str)
>              (match-string 0 str)))
>        (m* (when (string-match mod-regx* str)
>              (match-string 0 str))))
>   (lwarn 't :warning "m+=%s, m*=%s" m+ m*))
>
> I get
>
>   Warning (t): m+=C-S-, m*=
>
> Should not both this regexp give the same result here? It looks to me like
> * is treated as *? - ie non-greedy.

The regexp matches the empty string.

Andreas.

-- 
Andreas Schwab, SuSE Labs, schwab@suse.de
SuSE Linux Products GmbH, Maxfeldstraße 5, 90409 Nürnberg, Germany
PGP key fingerprint = 58CA 54C7 6D53 942B 1756  01D3 44D5 214B 8276 4ED5
"And now for something completely different."

Re: Should not * be greedy in a regexp?

[Andreas Schwab]

Andreas Schwab <schwab@suse.de> writes:

> "Lennart Borgman (gmail)" <lennart.borgman@gmail.com> writes:
>
>> Starting from
>>
>>    emacs -Q
>>
>> Put this in the *scratch* buffer and eval it:
>>
>> (let* ((mod-regx+ "\\(\\(?:[CSM]-\\)+\\)")
>>        (mod-regx* "\\(\\(?:[CSM]-\\)*\\)")
>>        (str "<C-S-tab>")
>>        (m+ (when (string-match mod-regx+ str)
>>              (match-string 0 str)))
>>        (m* (when (string-match mod-regx* str)
>>              (match-string 0 str))))
>>   (lwarn 't :warning "m+=%s, m*=%s" m+ m*))
>>
>> I get
>>
>>   Warning (t): m+=C-S-, m*=
>>
>> Should not both this regexp give the same result here? It looks to me like
>> * is treated as *? - ie non-greedy.
>
> The regexp matches the empty string.

... at the start of the string, I wanted to add.

Andreas.

-- 
Andreas Schwab, SuSE Labs, schwab@suse.de
SuSE Linux Products GmbH, Maxfeldstraße 5, 90409 Nürnberg, Germany
PGP key fingerprint = 58CA 54C7 6D53 942B 1756  01D3 44D5 214B 8276 4ED5
"And now for something completely different."

Re: Should not * be greedy in a regexp?

[Lennart Borgman (gmail)]

Andreas Schwab wrote:
> Andreas Schwab <schwab@suse.de> writes:
> 
>> "Lennart Borgman (gmail)" <lennart.borgman@gmail.com> writes:
>>
>>> Starting from
>>>
>>>    emacs -Q
>>>
>>> Put this in the *scratch* buffer and eval it:
>>>
>>> (let* ((mod-regx+ "\\(\\(?:[CSM]-\\)+\\)")
>>>        (mod-regx* "\\(\\(?:[CSM]-\\)*\\)")
>>>        (str "<C-S-tab>")
>>>        (m+ (when (string-match mod-regx+ str)
>>>              (match-string 0 str)))
>>>        (m* (when (string-match mod-regx* str)
>>>              (match-string 0 str))))
>>>   (lwarn 't :warning "m+=%s, m*=%s" m+ m*))
>>>
>>> I get
>>>
>>>   Warning (t): m+=C-S-, m*=
>>>
>>> Should not both this regexp give the same result here? It looks to me like
>>> * is treated as *? - ie non-greedy.
>> The regexp matches the empty string.
> 
> ... at the start of the string, I wanted to add.

Ah, thanks.

Re: Should not * be greedy in a regexp?

[Stefan Monnier]

> Should not both this regexp give the same result here? It looks to me like *
> is treated as *? - ie non-greedy.

The regexp-matching is greedy, so it matches the longest match, but the
regexp-search will return the leftmost (i.e. first) match, so mod-regx*
will match at the very first position (matching the mprty string),
whereas mod-regx+ will fail at that position so it will then succeed at
the second position where both mod-regx+ and mod-regx* would match the
same string.


        Stefan