Try:

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

}
[download]

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'.

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

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.

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";
}
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to circumvent this look at defined (like suggested by busunsl) and for hashes also interesting exists.

-- Hofmator

#...your code...
print "Hello $name,\n ";
$secretword = $words{$name} || 'groucho';
#....more code written by you...
[download]

__________________________________________________

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

mexnix.perlmonk.org

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