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

I'm looking at this code in a book: 
$maximum = &max(3, 5, 10, 4, 6);

sub max {
 my($max_so_far) = shift @_;
 foreach (@_) {
  if ($_ > $max_so_far) {
   $max_so_far = $_;
  }
 }
 $max_so_far;
}

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So line 4: "my($max_so_far) = shift @_;" is a "$max_so_far" list variable? If so why is it a list variable, just because it's in parenthesis? Could it just have been typed as "my $max_so_far" and just let it be a scalar?

And line 10: "$max_so_far", is that just there so it is returned as the return value?

If it wasn't there, would the return value be the last boolean comparison made in the "if" statement?

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Re: Need to check my understanding of this code...
by davidrw (Prior) on Sep 04, 2005 at 19:12 UTC
    line 4: yes, your assumption is correct -- $max_so_far is a scalar .. These two lines are equivalent: 
    my($max_so_far) = shift @_;
my $max_so_far = shift @_;

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    You can, however, declare several scalars at the same time, in which case you use the parens: 
    my ($x, $y, $x) = (1, 2, 3);

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    As for line 10, yes, that last line does the same as return $max_so_far; (which would have been more explicit). From perldoc -f perlsub: "The return value of a subroutine is the value of the last expression evaluated."
    Trying it with line 10 commented out, $maximum is undef .. i tried w/the max value last in the list, too, so i supsect it's not the last "if", but that the undef comes from the loop termination..
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[d/l]
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Re: Need to check my understanding of this code...
by borisz (Canon) on Sep 04, 2005 at 19:04 UTC
    line 4 gets the first entry from the list. It does the same as my $max_so_far = shift. And $max_so_far is the last statement and returned to the calling sub. So it is like return $max_so_far;
    Boris
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      Not quite. Shift removes the first entry from the list, a simple assignment doesn't (although since that list is never used again, in this context it makes no difference - except that the assignment is slightly faster). Compare:

      #!/usr/bin/perl
use strict;
use warnings;
my @list = qw/a b c/;
my ($x) = @list;
print "X: $x, List: @list\n";
my $y = shift @list;
print "Y: $y, List: @list\n";

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      The output being:

      X: a, List: a b c
Y: a, List: b c
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[d/l]
        ???, I think one of us misread the question. I talk about the difference from my $x = shift; and my ($x) = shift @_; sure, there is a difference, but not in the context of the question.
        Boris
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Re: Need to check my understanding of this code...
by merlyn (Sage) on Sep 04, 2005 at 19:23 UTC
    this code in a book
    My, that text looks very familiar. Are you using Learning Perl 3rd edition, or 4th edition?

    As for your questions... I thought we addressed both of those points fairly head on in the surrounding text. Are you reading the text and doing the exercises, or just skipping around wondering about things?

    -- Randal L. Schwartz, Perl hacker
    Be sure to read my standard disclaimer if this is a reply.

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      I'm using the 4th edition. I'm reading straight through but only get an hour here or there, I'd like to do 100% of the exercises but I just don't always have computer access when I'm reading.

      And you did address it in the surrounding text. I had to stop reading right after that exercise to change some diapers (not mine) and got most of my questions answered in the follow up paragraph.

      You're book is great, I bought some others that aren't 1/4 as good.

      I really like how you show 4 or 5 varying examples of what you're explaining.

      I think everyone learning Perl must read your book.
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