Am I going crazy with the Perl string comparison operator? Debug log included -

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I should remember something about variable assignment or string comparison. I have a script that is running through a different file from one tab. Unless there is a special value in the line "P", I go to the next line, the code looks like this: 

  1 print "processing inst_report file ... \ n"; 2 foreign (@inst_report_file) {3 @ line = partition (/ \ t /); 4 ($ line [13] "p") & amp; Amp; the upcoming; 5 $ inst_report {$ line [1]} ++; 6}   
 For some reason, the script will never be found on line 5, even if there are clearly lines with "P" in it.  
 Then debug time!  
  # continue the breakpoint of the DB & lt; 13 & gt; C Main :: (count.pl:27): ($ line [13] "P") & amp; Amp; the upcoming; Explaining that this particular array element is actually "P", in which there is no leading or trailing character. DB & lt; 13 & gt; P "- ​​$ line [13] - \ n"; - P - # Prove that I am not mad and the Perl string comparison operator actually works. DB & lt; 14 & gt; P ("p" ek "p"); 1 # Now we have shown that the line $ [13] ek p, let's go boolean again. DB & lt; 15 & gt; P ($ line [13] eq "p") # (empty means wrong) watt? # Come and set $ line manually [13] DB < 16 & gt; $ Line [13] = "P" # Now we try that comparison ... db < 17 & gt; P ($ line [13] eq "p") 1 db < 18 & gt; # Now it is working. Why?   
 I can work by prefining the input file around it, but bothers me why it does not work. Am I clarifying something?  
 --- Lorraine ---   
 
 
 Find out that your string is actually What is using:  
  Use the data: Dumper; Local $ Data :: Dumper :: Use CQ = 1; Print (dumper ($ line [13]));   
 
 [On the further review, the estimations given below are most incorrect.]  
 I suspect that you have a trailing new line, the situation where You want in  
 You may have an empty space behind. Both  s / \ s + \ z //  will remove the following spaces and subsequent Newline.

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