Re: how to run regex calculation

I am not subscribed to the list, so sending to the authors too.

This seem to work:

echo "2 3 6" |perl -Mstrict -pE 's{(\d+)\s(\d+)\s(\d+)}{my 
$r="false";$r="true" if $3 == eval"$1*$2";$r}e;'

perl -Mstrict -lE 'my $str = "2 3 7"; $str =~ s{(\d+)\s(\d+)\s(\d+)}{my 
$result="false";$result="true" if $3 == eval"$1*$2";$result}e;say $str'

The magic is the /e modifier.

Regards,
Alex

On 1/9/20 5:25 PM, MIchael Capone wrote:
> It probably won't ever work.  The problem is the * (star/asterisk) after 
> \1 is being interpreted in the context of regular expressions, ie, 
> "zero-or-more matches", and not as a multiplication operator.  In fact,
> 
> ~]$ perl -Mstrict -le 'my $str = "2 3 23"; print "true" if 
> $str=~/(\d+)\s(\d+)\s(\1*\2)/'
> 
> ...does indeed print "true"
> 
> It would probably make the most sense to not try to do this as a 
> one-liner (why do we perl programmers love to be cute like that?), and 
> simply break it up into two steps:
> 
> $str =~ s/(\d+)\s(\d+)\s(\d+)/;
> print "true" if ($1*$2 == $3);
> 
> - Michael
> 
> 
> 
> On 1/9/2020 12:39 AM, Wesley Peng wrote:
>> Hallo
>>
>> on 2020/1/9 16:35, demerphq wrote:
>>> $str=~/(\d+)\s(\d+)\s(\1*\2)/
>>>
>>> $1 refers to the capture buffers from the last completed match, \1
>>> inside of the pattern part of a regex refers to the capture buffer of
>>> the currently matching regex.
>>
>> This doesn't work too.
>>
>> perl -Mstrict -le 'my $str = "2 3 6"; print "true" if 
>> $str=~/(\d+)\s(\d+)\s(\1*\2)/'
>>
>> nothing printed.
>>
>>
>> Regards.