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regular expression problem

Posted on 2003-03-18
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Last Modified: 2010-03-05
$string1="1+";
$string2="2+3+4+";

while($string1=~/([^\+]+)/g)
  {$a=$1;
   print("a=$a\n");
   while($string2=~/([^\+]+)/g)
     {$b=$1;
      print("b=$b\n");
      &SomeMoreRegularExpressions($a,$b);
     }
  }

This code is supposed to print output:
a=1
b=2
b=3
b=4

But it prints
a=1
b=2
b=3
b=4
b=3
b=4
b=4
b=4

What is wrong? How can I fix it?
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Question by:serg111
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9 Comments
 
LVL 8

Expert Comment

by:jhurst
ID: 8163820
you are doing a while and not chabnging the condition tested in the while.  
0
 
LVL 3

Expert Comment

by:prady_21
ID: 8164086
I tested your code on my system and it is working perfect.
I think you should check the regular expressions after the
print statement.ie after  --print("b=$b\n");

i just did cut and paste and you can check the result


#!/usr/bin/perl

### a test program

$string1="1+";
$string2="2+3+4+5+";

while($string1=~/([^\+]+)/g)
 {
  $a=$1;
  print("a=$a\n");
  while($string2=~/([^\+]+)/g)
    {
     $b=$1;
     print("b=$b\n");
    }
 }





%./resttest.pl
a=1
b=2
b=3
b=4
b=5
0
 
LVL 84

Expert Comment

by:ozo
ID: 8164144
what does
&SomeMoreRegularExpressions($a,$b);
do?
0
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Expert Comment

by:desktop2
ID: 8164209
Problem is somewhere in this function SomeMoreRegularExpressions since without it the while with regular expression is working fine


sub SomeMoreRegularExpressions{
local($ip,$port)=@_;
local(@output,$output,$path);
$childID=fork();
if($childID)
  {return 0; }
if($PathWww=~/([^\/]+)(\/.+)/)
 {$ip=$1;
  $path=$2;
 }
else
 {$path="/";
 }
$socket=IO::Socket::INET->new(Proto=>"tcp",PeerAddr=>$ip,PeerPort=>$port,Reuse =>1,Timeout=>40);
$request="POST $path HTTP/1.0\nContent-Type: application/x-www-form-urlencoded\nContent-Length: 7\n\nr=test\n";
print $socket $request;
@output = <$socket>;
# ===============================
foreach(@output)
  {$output.=$_;}
close($socket);

if($output eq "OK" || ($output=~/\r?\n\r?\nOK/g))
 {return 1; }
else
 { exit; }
}
0
 

Expert Comment

by:desktop2
ID: 8164219
actually $string1 and $string2 contains names of servers and port number to connect
0
 
LVL 20

Accepted Solution

by:
jmcg earned 800 total points
ID: 8168380
If the code Desktop2 found in one of your other questions is still the code you're using, the explanation of the behavior you are seeing is pretty clear:

Each value found in string 2 results in an additional process. When the output of the request handled by the child is "OK", you end up with both processes continuing to process the list in string2. The first value gets processed by the first child. The original process and the first child both process the second value. And each of them creates a child so there end up being 4 processes working on the 3rd value.

The obvious fix is to change your subroutine so that the child process always exits.
0
 

Expert Comment

by:desktop2
ID: 8168446
Thanks, it helped!
mmm... can't find any button to accept this answer
0
 
LVL 2

Author Comment

by:serg111
ID: 8168461
Sorry - I've connected from home computer with different username
0
 
LVL 20

Expert Comment

by:jmcg
ID: 8168676
Ah, the mystery of how desktop2 knew what was in the subroutine is now solved.

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