On 7/15/10 Thu Jul 15, 2010 12:24 PM, "Chandan Kumar"
scribbled:
> Hi guys,
>
> I could able to understand the quantifiers. But still stuck with one
> confusion.if dot operator is used with quantifiers.
>
> ex: my string is $_="ths is my first regular expression";
>
> please explain t
Hi guys,
I could able to understand the quantifiers. But still stuck with one
confusion.if dot operator is used with quantifiers.
ex: my string is $_="ths is my first regular expression";
please explain the below questions with answer.
1)thi.*?
2)thi.+?
3)thi.+?s
4)thi.*?s
Thanks
Chas. Owens wrote:
On Thu, Jul 15, 2010 at 09:52, Dr.Ruud wrote:
perl -wle '
print --$| for 1..4;
'
1
0
1
0
snip
Interesting behavior. That isn't documented in perlvar.
see perlgolf
:)
--
Ruud
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On Thu, Jul 15, 2010 at 14:04, Shawn H Corey wrote:
snip
>> Yeah, I am an idiot. My only excuse is that I had just woken up.
snip
> I find it best not to answer questions before the first cup of coffee. ;)
snip
I haven't had any caffeine for over two years now. That may explain
some things.
-
On 10-07-15 12:47 PM, Chas. Owens wrote:
On Thu, Jul 15, 2010 at 11:50, Uri Guttman wrote:
"CO" == Chas Owens writes:
CO> No, you can say
CO> $a = !a;
CO> or
CO> $a != $a;
that is a no-op. ! isn't a binary op so it can't be used as an
assignment op. != is just the numeric com
> "CO" == Chas Owens writes:
CO> On Thu, Jul 15, 2010 at 11:50, Uri Guttman wrote:
>>> "CO" == Chas Owens writes:
>>
>> CO> No, you can say
>>
>> CO> $a = !a;
>>
>> CO> or
>>
>> CO> $a != $a;
>>
>> that is a no-op. ! isn't a binary op so it can't be used
On Thu, Jul 15, 2010 at 06:12, Sooraj S wrote:
snip
>> Opening a telnet connection does not mean that Perl code will run on
>> the remote system. You should also not be using telnet as it is very
>> insecure. Look up ssh, configure your ssh keys, and try something
snip
> Thanks for your reply.
On Thu, Jul 15, 2010 at 11:50, Uri Guttman wrote:
>> "CO" == Chas Owens writes:
>
> CO> No, you can say
>
> CO> $a = !a;
>
> CO> or
>
> CO> $a != $a;
>
> that is a no-op. ! isn't a binary op so it can't be used as an
> assignment op. != is just the numeric comparison not equal op. so that
> Hi,
> Thanks for your reply. Actually i need to login to a remote machine
> and do some series of steps including creating,editing and taring some
> files and executing some other scripts..The remote machine is in our
> local network itself..so there is no security issues and thats why i
> am us
> "CO" == Chas Owens writes:
CO> No, you can say
CO> $a = !a;
CO> or
CO> $a != $a;
that is a no-op. ! isn't a binary op so it can't be used as an
assignment op. != is just the numeric comparison not equal op. so that
last line is always false and doesn't change $a.
uri
--
Uri G
On Thu, Jul 15, 2010 at 09:52, Dr.Ruud wrote:
> Bryan R Harris wrote:
>
>> Out of curiosity, is there a unary not operator in perl?
>>
>> i.e. "$a = $a+1" is the same as "$a++"
>>
>> Is there a similarly short form of "$a = !$a"? Like "$a!!"? (tried it
>> and
>> it didn't work.)
>
> perl -
Bryan R Harris wrote:
Out of curiosity, is there a unary not operator in perl?
i.e. "$a = $a+1" is the same as "$a++"
Is there a similarly short form of "$a = !$a"? Like "$a!!"? (tried it and
it didn't work.)
If it was there it should be ^^, not !!.
$a += 1 <=> $a++
$a
On Jul 15, 4:57 am, chas.ow...@gmail.com ("Chas. Owens") wrote:
> On Wed, Jul 14, 2010 at 10:16, Sooraj S wrote:
> > Hi I am very new to perl. I want to login to a remote machine and
> > check a directory exists or not.
>
> > my code:
> >
>
> > using Net::Telnet;
>
> > $t = new Net::Teln
On Sat, Jul 10, 2010 at 11:13 PM, Jeff Pang wrote:
> Session or Cookie is a general way for this purpose.
> Are you programming under mod_perl?
> Even under mod_perl CGI::Session is better than Apache::Session from my
> experience.
> Or you could try CGI::Cookie instead.
Though a cookie is only
Bryan R Harris wrote:
Out of curiosity, is there a unary not operator in perl?
i.e. "$a = $a+1" is the same as "$a++"
Is there a similarly short form of "$a = !$a"? Like "$a!!"? (tried it and
it didn't work.)
perl -wle '
print --$| for 1..4;
'
1
0
1
0
--
Ruud
--
To unsubscribe,
On Thu, Jul 15, 2010 at 02:32, Bryan R Harris
wrote:
>
>
> Out of curiosity, is there a unary not operator in perl?
>
> i.e. "$a = $a+1" is the same as "$a++"
>
> Is there a similarly short form of "$a = !$a"? Like "$a!!"? (tried it and
> it didn't work.)
snip
No, you can say
$a = !a;
o
On Thu, Jul 15, 2010 at 03:22, Thomas Bätzler wrote:
snip
> In RE, a "." by itself is the atom that matches any one character.
>
> The quantifiers mean:
>
> "?" 0 or 1 occurences of the previous expression.
> "+" 1 or more occurrences of the previous expression (greedy match).
> "*" 0 or more occu
2010/7/15 Bryan R Harris :
>
>
> Out of curiosity, is there a unary not operator in perl?
>
> i.e. "$a = $a+1" is the same as "$a++"
>
> Is there a similarly short form of "$a = !$a"? Like "$a!!"? (tried it and
> it didn't work.)
>
I dont think that's helpful much for the actual programmin
Out of curiosity, is there a unary not operator in perl?
i.e. "$a = $a+1" is the same as "$a++"
Is there a similarly short form of "$a = !$a"? Like "$a!!"? (tried it and
it didn't work.)
- Bryan
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Hi ,
Thanks guys for your quick response.
I will try it out combinations and get back to you if i have issues.
Regards,
chandan.
--- On Thu, 15/7/10, Thomas Bätzler wrote:
From: Thomas Bätzler
Subject: AW: Query on Qunatifiers
To: "Beginners Perl"
Cc: "Chandan Kumar"
Date: Thursday
On Jul 14, 2:59 pm, chas.ow...@gmail.com ("Chas. Owens") wrote:
> On Wed, Jul 14, 2010 at 00:50, C.DeRykus wrote:
>
> snip>> s/(\w+)\s/$1 /g;
> snip
> > Neat. Using the \K construct available in 5.10, you can even
> > eliminate the need for the capture:
>
> > s/ \w+ \K \s / /gx;
>
> snip
>
>
Chandan Kumar asked:
> I have query over quantifiers.
>
> Could you please explain the combination of operators Question mark
> (?),dot(.),star(*),plus(+).
>
> Say this is my string:
>
> $_ = " this is my first pattern ,quite confused with quantifiers"
>
> ex: (thi.*?) or (thi.+?) etc
>
>
On Thursday 15 Jul 2010 09:51:32 Chandan Kumar wrote:
> Hi ,
>
> I have query over quantifiers.
>
> Could you please explain the combination of operators Question mark
> (?),dot(.),star(*),plus(+).
> Say this is my string:
>
> $_ = " this is my first pattern ,quite confused with quantifi
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