How to start regexp from a X position

pkt has asked for the wisdom of the Perl Monks concerning the following question:

Hello

I was tried to parse and order something like this

my $x = "<0>";

for (('<1>','<2>','<3>')) {
   $x =~ s/(<0>)/${1}${_}/;
}
[download]

so i get this output:

<0>
<3>
<2>
<1>
[download]

But i want to order from smaller to bigger

<0>
<1>
<2>
<3>
[download]

how can i do to change it using regexp option, I was saw c modifier and \G, but i'm not sure how to use that

my $x = "<0>";

for (('<1>','<2>','<3>')) {
   if ($x =~ /<\d>/gc) {
      $x =~ s/(\G<\d>)/${1}${_}/g;
   }
}
[download]

i was saw too pos function but it's a read only funcion or what's SCALAR mean in the documentation.

Thank a lot in advance

Nicolais

-----------------------------

Update: Basicaly i was tried to replace in the last match, how could i do that? using regexp only and of course it functions like pos, the best and good style way.

Comment on How to start regexp from a X position Select or Download Code

Replies are listed 'Best First'.
Re: How to start regexp from a X position by ikegami (Patriarch) on Oct 07, 2008 at 11:35 UTC
It's hard to tell what a good solution would be without knowing more about the real problem. I can think of two trivial solutions. Use `reverse` to process the items in the opposite order. `my $x = "<0>"; for (reverse '<1>', '<2>', '<3>') { $x =~ s/(<0>)/${1}${_}/; }` [download] Just do one replace. `my $x = "<0>"; $x =~ s/(<0>)/$1 . join('', '<1>', '<2>', '<3>')/e;` [download]	[reply] [d/l] [select]
Re^2: How to start regexp from a X position by pkt (Initiate) on Oct 07, 2008 at 12:09 UTC
Yes, many thank I wasn't clear about the question, basicaly i want replace in the last match, using only regexp and it's funcion like pos, in fact, i couldn't understand how pos function apart of course that retrieve the last position. Thank, ikegami	[reply]
Re^3: How to start regexp from a X position by ikegami (Patriarch) on Oct 07, 2008 at 12:11 UTC
basicaly i want replace in the last match Why? If you've just put it there, why would you like to proceed to replace it? using only regexp and it's funcion like pos Why? It's odd to tell someone how to solve a problem you can't solve.	[reply]
Re^4: How to start regexp from a X position by pkt (Initiate) on Oct 07, 2008 at 12:47 UTC
Re^4: How to start regexp from a X position by Anonymous Monk on Oct 07, 2008 at 13:02 UTC
Re: How to start regexp from a X position by JavaFan (Canon) on Oct 07, 2008 at 12:09 UTC
Here a way of doing it with pos: `my $x = "<0>"; my $pos = 0; for (qw {<1> <2> <3>}) { pos($x) = $pos; $x =~ s/\G(<\d+>)/$1$_/; $pos += length $1; } say $x; __END__ <0><1><2><3>` [download]	[reply] [d/l]
Re^2: How to start regexp from a X position by pkt (Initiate) on Oct 07, 2008 at 12:20 UTC
Thank!! great!!!, you open my mind, really i was knew that the solution is around that but i couldn't understand the documentation, perfect!!!!. One last question, how works the modifier c and \G in a regexp Could i do something like this: `my $x = "<0>"; for (qw {<1> <2> <3>}) { if ($x =~ /<\d+>/gc) { $x =~ s/\G(<\d+>)/$1$_/; } } say $x;` [download] Really thank JavaFan	[reply] [d/l]
Re^3: How to start regexp from a X position by JavaFan (Canon) on Oct 07, 2008 at 12:44 UTC
I've no idea how to incorporate the /c modifier. It'll be pointless for two reasons. First, one sets pos() before matching, so who cares about the effect of /c. Second, /c would only kick in if you add a substring `foo` to $x, and the next time around, `foo` isn't there.	[reply]
Re^4: How to start regexp from a X position by pkt (Initiate) on Oct 07, 2008 at 13:01 UTC
Re^3: How to start regexp from a X position by ikegami (Patriarch) on Oct 07, 2008 at 12:33 UTC
Thank!! great!!!, you open my mind, You picked the worst of the presented solutions solely because it used `pos` as you initially believed it should. That's the very definition of a closed mind.	[reply] [d/l]
Re^4: How to start regexp from a X position by pkt (Initiate) on Oct 07, 2008 at 12:57 UTC
Re: How to start regexp from a X position by JavaFan (Canon) on Oct 07, 2008 at 11:41 UTC
`pos` is not a read-only function, you can actually assign to it. However, that won't help you much, as you're matching `<0>`, and that only occurs once in the string. As for you example, it can be written as: `$x =~ /(<0>)/$1<1><2><3>/` [download]	[reply] [d/l] [select]
Re^2: How to start regexp from a X position by pkt (Initiate) on Oct 07, 2008 at 12:01 UTC
Thank for your reply Yes, i matching <0> in the first example, but i can change it to <\d>, could be it help me with pos? i'm not put the question cleary, i was tried to replace in the last match, how could i do that? Thank again	[reply]
Re^3: How to start regexp from a X position (5.10 regex feature) by moritz (Cardinal) on Oct 07, 2008 at 12:15 UTC
You can get the last match by adding `.` in front of the regex. If you use perl 5.10.0 or newer, you can make use of the `\K` ("keep") assertion: `use 5.010; my $str = '<1> <2> <3>'; $str =~ s/.\K<\d>/<4>/; say $str; __END__ <1> <2> <4>` [download] If you want to add test after the match, you can use `s/.*<\d>\K/ <4>/` instead.	[reply] [d/l] [select]
Re^4: How to start regexp from a X position (5.10 fregex feature) by pkt (Initiate) on Oct 07, 2008 at 12:32 UTC
Re^3: How to start regexp from a X position by ikegami (Patriarch) on Oct 07, 2008 at 12:08 UTC
`for ('<1>', '<2>', '<3>') { $x =~ /^(.<\d>)/$1$_/s; }` [download] is just a complicated way of doing `$x =~ /^(.<\d>)/$1<1><2><3>/s;` [download]	[reply] [d/l] [select]
Re^4: How to start regexp from a X position by pkt (Initiate) on Oct 07, 2008 at 12:27 UTC
Re^5: How to start regexp from a X position by ikegami (Patriarch) on Oct 07, 2008 at 12:42 UTC