Scoping Problem

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

I thought I had gotten a pretty good grasp of scoping and variables both from the various tutorials here, several classes on Perl and also the Camel Book. I was testing out a section of code that will manipulate a variable within a code block and I was thinking that I would not have to reassign it back to the original value when I was finished since I assign it once at the start of the main package. I proceeded to test this out with the following code snippet.

#! /usr/local/bin/perl -w

use strict;

my $hero="Batman";
my $car="Pinto";
my $year=sprintf("%04d", (localtime)[5]+1900);

BLOCK0:{ print "In the first block, \$hero is $hero and \$car is $car.
+\n"; }
BLOCK1:{ my $hero="Superman";my $car="Porsche";print "In next code blo
+ck \$hero is $hero and \$car is $car.\n"; }
BLOCK2:{ print "In next code block, \$hero is $hero and  \$car is $car
+.\n"; }
BLOCK3:{ print "To start, \$year is $year\n";}
BLOCK4:{ my $year=--$year; print "We change \$year to $year\n"; }
BLOCK5:{ print "And now, \$year is $year\n"; }

__DATA__

In the first block, $hero is Batman and $car is Pinto.
In next code block $hero is Superman and $car is Porsche.
In next code block, $hero is Batman and  $car is Pinto.
To start, $year is 2003
We change $year to 2002
And now, $year is 2002
[download]

According to everything I've seen, the last line should be: And now, $year is <b>2003</b>. Obviously, this is not the case and I am trying to figure out why this is so. I have also tried hardcoding a value into $year at the start and still get the same result. I am suspecting the decrement operator(?) is doing something to the value of $year in the main package, but what and how.

"Ex libris un peut de tout"

Comment on Scoping Problem Select or Download Code

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Re: Scoping Problem by VSarkiss (Monsignor) on Jun 25, 2003 at 18:55 UTC
In `my $year = --$year;`the decrement operator is changing the `$year` in the OUTSIDE scope, then assigning it to the lexical `$year` in the INSIDE scope (which is the one that your print statement in `BLOCK4` sees). Remember, until that statement finishes successfully, the INNER `$year` doesn't exist. You could get the behavior you're looking for with: `my $year = $year - 1;`which doesn't change the value of the OUTER `$year`. Update Added example in second para.	[reply] [d/l] [select]
Re: Re: Scoping Problem by nimdokk (Vicar) on Jun 25, 2003 at 19:02 UTC
Thanks, I actually tried it with `my $year=$year-1;` and not seen any problems. I think I'll just have to stick with that. "Ex libris un peut de tout"	[reply] [d/l]
Re: Scoping Problem by tcf22 (Priest) on Jun 25, 2003 at 19:04 UTC
`--$year` actually decrements the variable `$year` outside the block, because the right side is evaluated first, before the local copy is created. It doesn't simply return the value 1 less than year and assign it to the local variable year.	[reply] [d/l] [select]
Re: Scoping Problem by antirice (Priest) on Jun 25, 2003 at 20:42 UTC
`BLOCK4:{ my $year=$year;--$year; print "We change \$year to $year\n"; +}` [download] As mentioned before, $year's scope doesn't change until after `my $year = $year;` has executed. Thus, if you change $year anywhere after that statement executes, it will just change the $year within the scope and leave global $year alone. antirice The first rule of Perl club is - use Perl The ith rule of Perl club is - follow rule i - 1 for i > 1	[reply] [d/l] [select]