The real issue you have is that when the hash is empty, @keyarr gets no values from keys %hash. Thus, $keyarr[0] is undefined, causing the warning.

However, when the hash is not empty, @keyarr has values, so $keyarr[0] is defined and doesn't trigger the warning.

Add a print statement to show what's in @keyarr and then it may be clearer to you what is happening.

my (%hash);
#my (%hash) = ( foo => 'bar' ); # Try this variation
my (@keyarr)=(keys (%hash));

print "keys: '@keyarr'\n";      # Add this line

if (defined ($hash{$keyarr[0]}))
{print "hash is defined\n";}

else
{print "hash is undefined\n";}
[download]

xdg, you're right, but then the question is why one would use a variable that is potentially undefined??
for this, testing whether the hash has values is better like in muntfish's solution.
But if you go that route, why don't you directly test if (scalar(%hash)!=0) { ....??
using a hash in scalar context gives 0 for a empty hash and the number of used entries for a nonempty hash.
my %hash = (1=>2,3=>4);print "scalarval=".scalar(%hash); => 2/8

Cheers,
Rob

xdg, you're right, but...

I didn't recommending that approach. I just answered the original question. Sometimes a limited answer is the one that people need in order to make their own discoveries. (See "Baby" Perl versus "Bad" Perl.)

-xdg

Code written by xdg and posted on PerlMonks is public domain. It is provided as is with no warranties, express or implied, of any kind. Posted code may not have been tested. Use of posted code is at your own risk.

This is harder to do than you think. If you run with use strict; somewhere in your code then the program will not run unless %hash is defined.

If you run the code without use strict; somewhere in your code then the program will run whether %hash is declared or not. But there lies the next problem: the hash will magically be created the very first time you refer to it. Therefore the mere fact you are trying to access the hash means that it must exist.

So the only problem you can solve is whether the hash contains anything. If the hash is empty then (@keyarr == 1). Otherwise (@keyarr != 1).

Update: as muntfish correctly identified (@keyarr == 0) is the test to use.

To clarify - unlike scalars, arrays and hashes can't be "undefined". They can be empty, which is slightly different.

monarch, to test for emptiness, you said:

If the hash is empty then (@keyarr == 1).

I think you mean 0 not 1:

my %hash;
my @keyarr = keys %hash;
print scalar @keyarr;
[download]

prints 0.

---

s^^unp(;75N=&9I<V@`ack(u,^;s|\(.+\`|"$`$'\"$&\"\)"|ee;/m.+h/&&print$&

#!/usr/bin/perl

use strict;
use warnings;

my %hash;
print %hash ? 'defined' : 'undefined';
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You're actually not checking to see if %hash is defined, but only if that specific element of the hash is defined. This is clearly not the same thing.

Now, as the prior respondents (all of whom have more Perl-dharma than do I) have noted, there are other ways of seeing if the hash is defined.

print "\%hash is defined\n" if(%hash);
[download]

or

you could try something like

$defined = keys(%hash);
[download]

My preference is for the first; it looks like it involves less work for Perl. The second looks like it's extracting all the keys so it can count them, and then throws away everything except the count. This seems wasteful.