[prog] quick perl question

[Caroline Johnston]

Many thanks. 

Is it ok to declare a variable in a loop? Does it get binned at the end 
and recreated at the start of every loop? Does it matter if it does?

Cxx


On Fri, 5 Mar 2004, Jacinta Richardson wrote:

> 
> On Thu, 4 Mar 2004, Caroline Johnston wrote:
> 
> >    {
> >       my $rowlength = 0;
> >       my $pos= 0;
> >       foreach my $row (@rows) 
> >       {
> >          next if $row =~ /^\s$/;
> >          push @matrix, [split /$delim/, $row];
> >          $rowlength = @{@matrix[$pos]} if ($rowlength == 0);
> > 	 die("different row lengths in data matrix") unless ($rowlength == @{@matrix[$pos]})
> >       
> >       }
> >    }
> > 
> > does that look OK? Better to use a for loop?
> 
> Nah.  Just changing one variable you avoid needing to know where you are
> in the list at all.
> 
> 
> >    {
> >       my $rowlength = 0;
> >       foreach my $row (@rows) 
> >       {
> >          next if $row =~ /^\s$/;
> 	   my @split = split /$delim/, $row];
> >          push @matrix, [@split];
> >          $rowlength = @split if ($rowlength == 0);
> >  	   die("different row lengths in data matrix") unless ($rowlength == @split})
> >       
> >       }
> >    }
> 
> Of course, there's more than one way to do it, and if you're someone who
> really, really doesn't like extra variables hanging around, you might
> prefer this answer:  (I prefer the former because it's more clear, but
> not by much)
> 
> >    {
> >       my $rowlength = 0;
> >       foreach my $row (@rows) 
> >       {
> >          next if $row =~ /^\s$/;
> >          push @matrix, [split /$delim/, $row];
> >          $rowlength = @{@matrix[-1]} if ($rowlength == 0);
> > 	 die("different row lengths in data matrix") unless ($rowlength == @{@matrix[-1]})
> >       
> >       }
> >    }
> 
> That's right.  Perl lets you count backwards through your arrays too.
> -1 is the last entry, -2 is the second last entry and so forth.
> (- at array) should give you position 0.  :)
> 
> > numify? really? Fantastic word!
> 
> Yup.  It goes next to stringify.  Which is what happens when you treat a
> scalar as a string.  
> 
> I hope this helps,
> 
> 	Jacinta
> 	
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