in reply to Re: if modifier of a print statement
in thread if modifier of a print statement

   /no_proc/ ? print "$_\n"
             : print "$_ : in proc\n" ;

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You are using the Conditional Operator in void context.   That is usually written as: 

   print /no_proc/ ? "$_\n" : "$_ : in proc\n";

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Re^3: if modifier of a print statement
by Marshall (Canon) on Jun 30, 2009 at 12:34 UTC
    I think what I have is correct. It compiles and runs under "use strict;" and "use warnings;".

    The /no_proc/ is a match statementoperator on $_ which yields a true/false value. If the line in $_ has "no_proc", it gets printed and that's it. If that line doesn't have "no_proc" then the 2nd "else" statement is printed...."$_..in proc".

    print will return a "true value". What would "print /adsf/;" mean? I think nothing, but I've been wrong before!

    Updated: changed "statement" to "operator"

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      I did not say that your code was "incorrect".   What I did say was that  ? : is usually used as an rvalue expression:

      my $variable = /condition/ ? 'TRUE' : 'FALSE';

      or as an lvaue expression:

      ( /condition/ ? $true : $false ) = 'something';

      but very rarely, if ever, used in a void context.

      The /no_proc/ is a match statement on $_ which yields a true/false value.

      The /no_proc/ is a match operator on $_ which yields a true/false value.

      print will return a "true value".

      It will if it worked correctly, otherwise it will return a "false value".

      What would "print /adsf/;" mean? I think nothing,

      "print /adsf/;" would print whatever /adsf/ returned, usually either 1 or '' for "true" or "false".   But that is not what the expression I wrote would do which would print either "$_\n" or "$_ : in proc\n" depending on whether /no_proc/ was "true" or "false".

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        I would respectfully disagree on some points and hopefully we can agree on the main point.

        I dunno what got my brain working on the ternary operator idea, maybe it was the "and" or "&&" stuff? (no pun intended). I normally don't use this syntax in Perl code.

        The main point here is that this is a simple "if","else" condition, not a more complex construction with "and" and "next". That's it! That concept is stunning in its simplicity. 

         # I think the if/else is better and I think
 # we agree on that. The if/else can be written like this, but
 # nobody cares! Let's just stop talking about it!
 # It simply does NOT matter. We are beating a dead horse if
 # we worry about this:
 #  /no_proc/ ? print "$_\n" 
 #            : print "$_ : in proc\n" ;  
 #
 
 # some very few extra parens results in this:
 # It is crystal clear, there are no "ands, nexts"
 # that is the MAIN point!
 #
 if (/no_proc/)
 {
     print "$_\n";
 }
 else
 {
     print "$_ : in proc\n" ;  
 }

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Re^3: if modifier of a print statement
by JavaFan (Canon) on Jul 01, 2009 at 09:58 UTC
    You can only write it that way because both clauses in the expression happen to call the same function. If the statement was: 
    /no_proc/ ? foo "$_\n"
          : bar "$_ : in proc\n";

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    you could not have applied such a transformation, and the ?: would be left in void context.

    But I don't understand the fear of void context of particular functions or operators. Some people claim to get utterly confused when encountering a map in void context, breaking their tiny little brains on the question whether the author might have intended something else. You don't seem to like ?: in void context. But then you happily use print in void context, totally ignoring its return value.

    Me, I happily use anything in void context if that suits me. And I ain't afraid in the dark either.

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