All0uette has asked for the wisdom of the Perl Monks concerning the following question:

As a two day learner of perl I've come across the following, basically I'm messing around with the examples in the O'reilly Learning Perl. Use of uninitialized value at perky-array line 38, <> chunk 1. The script still works but I get the following error when the 'loop' related to an unkown user kicks in ? Can anyone help an (obvious) initiate: 
#!/usr/bin/perl -w

###@words = ("camel", "llama", "alpaca");
###
###  #or
###  #@words = qw(camel llama alpaca); # qw acts to specify "" to each
+ variable
###

### Above shows the use of an 'array' - the following describes an 'ha
+sh' 
### shown as %a (see 'pinky' for full program)

%words = qw     (
                fred    camel
                barney  llama
                betty   alpaca
                wilma   alpaca
                );

print "What is your name? ";
$name = <>;
chomp $name;

if
  ($name eq "pete")
  {
     print "Hello, $name! \n";
  }
else
  {
     print "Hello $name,\n ";
     $secretword = $words{$name};            # gets secret word

####    problem seems to start here #####

     if  ($secretword eq "")
        {$secretword = "groucho"; }
     print "What's the secret word ?\n";

     $guess = <>;
     chomp ($guess);

     while ($guess ne $secretword)
     {
       print " Niet - try again, What is the secret word ?\n";
       $guess = <>;
       chomp ($guess);

     }
  }

[download]

Edit Masem 2001-08-22 - Code reformatting to make easier to read

Replies are listed 'Best First'.
Re: Uninitialized value ?
by busunsl (Vicar) on Aug 20, 2001 at 17:22 UTC
    When you try to get a value for a key which is not in the hash, you'll get an undefined value.
    That undefined value is not equal to "".
    You can check undefined values with defined.

    Try:

    $secretword = $words{$name};            # gets secret word
if (! defined $secretword) {

}

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      You fill your hash: 
      %words = qw     (

                fred    camel
                barney  llama
                betty   alpaca
                wilma   alpaca

                );

      [download]
      It now contains four elements with the names of famous actors as keys.

      If the user enters one of the keys, everything will be fine, because e.g. $words{fred} is defined, so $secretword will get the value 'camel'.
      If the user enters something, that is not in the list of keys, e.g. 'busunsl', then $words{busunsl} is not defined.
      And $secretword will become undef.

      And you check for an undef value with defined.

      You wanted to fill something into $secretword if there was nothing in it, that is, if it is undef.
      So your if-statement has to be: if (! defined $secretword) { With the ! meaning 'not'.

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        This however alters the way the program works. 
        if (!defined $secretword or $secretword eq '') {

        [download]
        It will now both check if $secretword has some value (possible empty) and if it is empty.
        This can be boiled down to the line 
        if (!$secretword) {

        [download]
        or if you like the unless operator: 
        unless ($secretword) {

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        T I M T O W T D I
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      busunsl, Thanks for that - it seems to work - although I'm still not sure why the original didn't. Especially as I got this form the book ! Any chance you can explain a lttle more. Cheers.
[reply]

        I think your code should work correctly - in both versions. The only difference is that you get once a warning (which is exactly that, it's not an error) and iirc these early examples in merlyn's book (I suspect you are reading the llama) don't run under -w and strict because he hasn't introduced all the concepts yet.

        In if ($words{$name} eq "") the key $name is looked up in the hash %words. This returns undef when the key $name does not exist in %words. The undef is then compared to "" with a string compare, so the undef is transformed into the empty string. This results in the comparision being true. But this might be a source of error, so perl warns you if you have warnings enabled (-w).

        Why might that be a source of error? Well, imagine the empty string would be a valid entry in your hash, then you get the same result no matter if the key does not exist or the value is empty. Some code: 

        #/usr/bin/perl -w
use strict;

my %hash;
$hash{key1} = "";

# this produces no warning
if ($hash{key1} eq "") { # true
  print "key1 empty!\n";
}

# this produces a warning
if ($hash{key2} eq "") { # true
  print "key2 empty!\n";
}

        [download]
        to circumvent this look at defined (like suggested by busunsl) and for hashes also interesting exists.

        -- Hofmator

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Re: Uninitialized value ?
by mexnix (Pilgrim) on Aug 20, 2001 at 18:00 UTC
    Try doing this: 
    #...your code...
print "Hello $name,\n ";
$secretword = $words{$name} || 'groucho';
#....more code written by you...

    [download]
    what this does is, if the name $name is not in your list, mexnix currently isn't, it automatically assigns "groucho" to $secretword. || is logical OR and you can find it in your perldoc perlop manpage. With this, you can even skip the if ($secretword eq "") { $secretword = "groucho"} stuff, because it already does it for you. My $0.02

    __________________________________________________

    s mmgfbs nf, nfyojy m,tr yb-zya-zy,s zfzphz,print;
- thanks japhy :)

    mexnix.perlmonk.org

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      and what this also does is bite you back horribly ;-)
      if your $words{$name} evaluates to false, i.e. '' or 0. So better - unless you can outrule those cases: $secretword = defined($words{$name}) ? $words{$name} : 'groucho';

      -- Hofmator

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