what's wrong with this?

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

Hello monks!

I have the following code :

use strict;
use warnings;
use Tie::File;

my $first   = shift || die "first file\n";
my $sec     = shift || die "second\n";

tie my @first,'Tie::File',$first or die "tie error $!\n";
tie my @sec  ,'Tie::File',$sec or die "tie error $!\n";

my (%h1,%h2,%common);

print "processing $first\n";
map {
  if(/^\s*(\?.*?)\s/) {
    $h1{$1}++;
  }
} @first;

print "processing $sec\n";
map {
  if(/^\s*(\?.*?)\s/) {
    $h2{$1}++;
  }
} @sec;

print "finding common keys\n";
map {
  if(exists $h2{$_}) {
    $common{$_}++;
  }
} keys %h1;

print "there are @{[length keys %common]}:\n";
system("pause");
for my $key(keys %common) {
    print "\n\n\n$key\n\n\n";
}

untie @first;
untie @sec;
[download]

I am using this to find common stuff inside 2 files . My question is :
- if I print length keys %common ,it outputs 3
- if I assign keys %common to an array , it's size is 159 . Why ?
Comment on what's wrong with this?
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Re: what's wrong with this? by ccn (Vicar) on Oct 24, 2008 at 08:23 UTC
length returns the length in characters of given argument keys in scalar context (your case) returns numner of keys ina hash So if your hash has 153 keys than `length '153'` returns 3 i.e. three characters change you code as: `print "there are @{[scalar keys %common]}:\n";` [download]	[reply] [d/l] [select]
Re: what's wrong with this? by jwkrahn (Abbot) on Oct 24, 2008 at 12:53 UTC
*what's wrong with this?* `my $first = shift \|\| die "first file\n"; my $sec = shift \|\| die "second\n";` [download] The `\|\|` operator has higher precedence than the `=` operator so you need to either enclose the assignment in parentheses or use the lower precedence `or` operator. You can accomplish the same result more simply with a single hash: #!/usr/bin/perl use strict; use warnings; my $first = shift or die "first file\n"; my $sec = shift or die "second\n"; open my $FIRST, '<', $first or die "$first: $!"; open my $SEC, '<', $sec or die "$sec: $!"; print "processing $first\n"; my %common = map { /^\s(\?.?)\s/, 0 } <$FIRST>; print "processing $sec\n"; /^\s(\?.?)\s/ and $common{ $1 }++ while <$SEC>; print "finding common keys\n"; delete @common{ grep !$common{ $_ }, keys %common }; print "there are @{[scalar keys %common]}:\n"; system 'pause'; for my $key ( keys %common ) { print "\n\n\n$key\n\n\n"; } [download]	[reply] [d/l] [select]
Re^2: what's wrong with this? by Anonymous Monk on Oct 24, 2008 at 14:21 UTC
The \|\| operator has higher precedence than the = operator so you need to either enclose the assignment in parentheses or use the lower precedence or operator. No cigar for you :) `C:\>perl -MO=Deparse,-p -e" my $foo = shift \|\| die 666;" (my $foo = (shift(@ARGV) \|\| die(666))); -e syntax OK C:\>perl -MO=Deparse,-p -e" my $foo = shift or die 666;" ((my $foo = shift(@ARGV)) or die(666)); -e syntax OK C:\>perl -e" my $foo = shift \|\| die 666;" 666 at -e line 1. C:\>perl -e" my $foo = shift or die 666;" 666 at -e line 1. C:\>` [download] Perl Idioms Explained - && and \|\| "Short Circuit" operators To \|\| or not to or and why. \|\| die or or die :) "or" or "\|\|" problems or vs \|\| precedence issue	[reply] [d/l]
Re: what's wrong with this? by Alien (Monk) on Oct 24, 2008 at 08:21 UTC
never mind. I'm stupid :)	[reply]
Re^2: what's wrong with this? by Anonymous Monk on Oct 24, 2008 at 10:02 UTC
You crashed at Roswell didn't you? :D	[reply]