How would I print the last element of a loop?

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

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Re: Looping by tachyon (Chancellor) on Apr 02, 2004 at 16:01 UTC
Sounds like homework. `print $array[-1]; my $last; for ( @array ) { $last = $_ }; print $last; my $big = -999999999; while(<FILE>) { chomp; $big = $_ if $_ > $big }; print $big;` [download] cheers tachyon	[reply] [d/l]
Re: Looping by Limbic~Region (Chancellor) on Apr 02, 2004 at 16:05 UTC
Anonymous Monk, How would I print the last element of a loop? `print "$array[-1]\n"; # If you mean last element of array my $last = keys %hash; my $counter = 0; for my $key ( keys %hash ) { $counter++; print "$hash{$key}\n" if $counter == $last; } # If you mean determine if this is last item in loop` [download] If I run a code and go trough a file and found 5 numbers hor can I print the larger number? `use Scalar::Util 1.10 'looks_like_number'; use List::Util 'max'; my @numbers; while ( <FILE> ) { chomp; push @numbers , $_ if looks_like_number( $_ ); } print max(@numbers), "\n";` [download] I would highly recommend you take a look at How do I post a question effectively?. You can find it and other helpful information at the PerlMonks FAQ. Cheers - L~R	[reply] [d/l] [select]
Re: Looping by Roy Johnson (Monsignor) on Apr 02, 2004 at 15:59 UTC
Typically, you would use a variable to keep track of the largest, as you read through the file. If you want to print only the last line of a file, you could do `while (<>) { print if eof }` [download] The PerlMonk `tr///` Advocate	[reply] [d/l]
Re: How would I print the last element of a loop? by Anonymous Monk on Apr 02, 2004 at 18:24 UTC
If this is homework and "spaceship operator" is allowed...to get the larger number: `@nums = (57, 113, 25, 76, 712, 13, 3); foreach (sort {$a <=> $b} @nums) { $larger = $_ } print "The numbers: @nums\n\nlarger=$larger\n";` [download]	[reply] [d/l]
Re: Re: How would I print the last element of a loop? by Fletch (Bishop) on Apr 02, 2004 at 19:00 UTC
Erm, you've combined an O(N log N) sort with an O(N) iteration (not to mention always performing N assignemnts); whereas a simple iterative scan will always be just O(N) and will at worst perform N assignments on an already sorted list. Not exactly the most efficient way to find it . . . Update: . . . then again maybe we need more answers like this to discourage homework questions. :)	[reply]