In Perl, what's the difference between "if defined $count" and "if $count"? - perl

Question

I have the following script:
#!/usr/bin/perl
use warnings;
use strict;
my $count = 0;
my ( #first , #second , #third );
while ($count <= 7){
push ( #first , $count);
push ( #second , $count) if defined $count;
push ( #third , $count) if $count;
$count++;
}
print "first: #first\n";
print "second: #second\n";
print "third: #third\n";
This produces the following output:
first: 0 1 2 3 4 5 6 7
second: 0 1 2 3 4 5 6 7
third: 1 2 3 4 5 6 7
What's the difference between putting if defined $count vs. if $count, other than the latter method won't add the zero to the array? I've searched the perldocs but couldn't find the answer.

Answer 1

Truth and Falsehood in perlsyn explains what values are considered false in a boolean context:
The number 0, the strings '0' and '',
the empty list (), and undef are
all false in a boolean context. All other values are true.
undef is the value of a variable that has never been initialized (or that has been reset using the undef function). The defined function returns true if the value of the expression is not undef.
if $count is false if $count is the number 0, the string '0', the empty string, undef, or an object that has been overloaded to return one of those things when used in a boolean context. Otherwise, it's true. (The empty list can't be stored in a scalar variable.)
if defined $count is false only if $count is undef.

Answer 2

if you see the documentation of defined in perldoc then you will find that
Returns a Boolean value telling
whether EXPR has a value other
than the undefined value undef. If EXPR is not present, $_ is
checked.
A simple Boolean test will not
distinguish among undef, zero, the
empty string, and "0" , which are all
equally false.
that means,
push ( #second , 'undef') if defined $count;
when $count = 0, then it is defined because 0 is different from undef and defined returns true, but in this case push ( #third , 'undef') if $count; if condition fails, that's why it is not pushing 0 into the array.

Answer 3

The defined predicate tests to see whether the variable ($count in this case) is defined at all. In the code you've written, it always will be defined inside the loop because it's always got some value.
If you were to add:
undef $count;
push ( #first , 'undef');
push ( #second , 'undef') if defined $count;
push ( #third , 'undef') if $count;
after the loop, you would see the difference. (Note that I changed the code to add the literal 'undef' instead of $count because you'll get misleading effects from adding the actual undef value.

Answer 4

The if decides to run its block (or single statement) by looking at the value of the expression you give it:
if( EXPR ) { ... }
If that expression is true, it runs its block. If that expression is false, it doesn't.
That expression can be just about anything. Perl evaluates the expression, reducing it to a value that is either true or false. The if() then looks at that value.
So, removing that part of your question, you're left with "What's the difference between defined $count and $count". Well, one is the return value for defined and the other is whatever value is stored in $count.
When you want to figure out what a particular bit of code is doing, reduce it in the same logical process that perl would, one step at a time. See what each step does, and you'll often be able to answer your own questions. :)
You say that you searched the documentation, but I'm not sure where you looked. If you want to look up a built-in function, you can use perldoc's -f switch:
$ perldoc -f defined
If you want to read about Perl's syntax for things such as if, that's in perlsyn.
I have a beginner's guide to the Perl docs in Perl documentation documentation.

Answer 5

The way I read it.
if $count is only true when $count evaluates != 0, hence the third array has no 0 in it.
if defined $count checks to see if the $count as a scalar has been created, and as you have $count scalar, it's the same as the first one.