Re: if modifier of a print statement

Yes, this is an op precedence problem. But why even confront that?

my @test = qw(no_proc line_to_proc1 line_to_proc2);
foreach (@test) 
{
   # this is actually just one line, indented to
   # make the if/else more clear. The trailing ";"
   # isn't even necessary. 
   # this looks like a simple "we do either a or b,
   # but we will do one of them for sure situation,
   # ie if/else. The single statement below probably
   # equates to exactly the same machine code as a
   # more lengthy if/else statement. There is no
   # "next", there is no "and".
   
   /no_proc/ ? print "$_\n" 
             : print "$_ : in proc\n" ;  
}

__END__
prints:
no_proc
line_to_proc1 : in proc
line_to_proc2 : in proc
[download]

Comment on Re: if modifier of a print statement Download Code

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Re^2: if modifier of a print statement by jwkrahn (Abbot) on Jun 30, 2009 at 12:14 UTC
`/no_proc/ ? print "$_\n" : print "$_ : in proc\n" ;` [download] You are using the Conditional Operator in void context. That is usually written as: `print /no_proc/ ? "$_\n" : "$_ : in proc\n";` [download]	[reply] [d/l] [select]
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* ~~statement~~operator 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"	[reply]
Re^4: if modifier of a print statement by jwkrahn (Abbot) on Jun 30, 2009 at 13:38 UTC
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 `"$_ : in proc\n"` depending on whether `/no_proc/` was "true" or "false".	[reply] [d/l] [select]
Re^5: if modifier of a print statement by Marshall (Canon) on Jun 30, 2009 at 15:58 UTC
Re^3: if modifier of a print statement by JavaFan (Canon) on Jul 01, 2009 at 09:58 UTC
You can only write it that way because both clauses in the expression happen to call the same function. If the statement was: `/no_proc/ ? foo "$_\n" : bar "$_ : in proc\n";` [download] 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.	[reply] [d/l]