Variable Declaration

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

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Re: Variable Declaration by Joost (Canon) on Aug 31, 2007 at 14:50 UTC
You know what would help? If you post some code that can be run and demonstrates the problem, including the complete error message. By vaguely describing the problem from memory, you're just making it hard for us to see what the problem is, and therefore you're making the chances of you getting a correct answer a lot smaller. Please read how (not) to ask a question, and writeup formatting tips. As in: use `<code>` tags for your code. "What should it profit a man, if he should win a flame war, yet lose his cool?"	[reply] [d/l]
Re^2: Variable Declaration by pp (Novice) on Aug 31, 2007 at 16:00 UTC
Here is the code: `use File::Basename; $infile_name = shift(@ARGV); if ( ! -r $infile_name) { print STDERR "1.Cannot read $infile_name\n"; exit 99; }` [download] a few more lines in between and in the main program I have the following: my $ver_name = shift @ARGV; my $mon_dir = dirname($infile_name); # Get directory if (! -z $ver_name) { (our $graph_dir = $mon_dir) =~ s?log/.$?abinitio/run?; #Build dire +ctory as above if not equal to version.(Replace log/ with abinitio/r +un) } else { (our $graph_dir = $mon_dir)=~ s?log/.$?$ver_name/abinitio/run?; #Build directory as such if equal to verion(Replace log/ with $ver_na +me/abinitio/run) } my $graph = "$graph_dir" . "/" . "$graphname"; # Concatenate graph directory with graphname if ( ! -r $graph) { print STDERR "2. Cannot read graph=$graph\n" exit 99; } [download] By including the if-then-else condition I am getting the following error that it "Cannot read graph=********(pathname)	[reply] [d/l] [select]
Re^3: Variable Declaration by ikegami (Patriarch) on Aug 31, 2007 at 16:19 UTC
A second error you haven't noticed yet: `$?` is a variable name. Perhaps you wanted `\$?`, although `.\$?` is the same as just `.`.	[reply] [d/l] [select]
Re: Variable Declaration by derby (Abbot) on Aug 31, 2007 at 14:51 UTC
Please use code tags. I think you want (not valid perl): `my $var2; if <cond1> { $var2 = a; } else { $var2 = y; }` [download] -derby	[reply] [d/l]
Re^2: Variable Declaration by ikegami (Patriarch) on Aug 31, 2007 at 15:09 UTC
The following would also work: `my $var2 = <cond1> ? a : y;` [download] But be careful. Even slight complexity in its arguments can make it hard to read.	[reply] [d/l]
Re^3: Variable Declaration by nobull (Friar) on Aug 31, 2007 at 16:11 UTC
A third option is `my $var2 = do { if ( <cond1> ) { a; } else { y; } };` [download] All three constructs are more-or-less interchangable and which one you use should be decided based on which is most readable in any particular instance.	[reply] [d/l]
Re: Variable Declaration by Roy Johnson (Monsignor) on Aug 31, 2007 at 15:09 UTC
`$var1` is defined in the scope of the `if`-blocks. It is not available outside them. Your example here is probably better written as `my $var2 = (cond 1) ? 'a' : 'y';` [download] Downvotes without explanation continue to be the scourge of Perlmonks. Caution: Contents may have been coded under pressure.	[reply] [d/l] [select]
Re: Variable Declaration by Anonymous Monk on Aug 31, 2007 at 15:06 UTC
The problem was solved by others before I got a crack at it but I couldn't resist explaining. It's not often I see something on here I could solve. You've got a classic beginner problem. This is scoping. Since you declared the variables (using "my") inside the if statement it only exists inside the if statement. So as soon as you leave your if statement you no longer have a value in $var1 so basically you assign nothing to $var2. Get Learning Perl and a big package of little yellow sticky notes to flag your favorite pages. gj	[reply]