The undef Value (Learning Perl, 3rd Edition)

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2.11. The undef Value

What happens if you use a scalar variable before you give it a value? Nothing serious, and definitely nothing fatal. Variables have the special undef value before they are first assigned, which is just Perl's way of saying "nothing here to look at -- move along, move along." If you try to use this "nothing" as a "numeric something," it acts like 0. If you try to use it as a "string something," it acts like the empty string. But undef is neither a number nor a string; it's an entirely separate kind of scalar value.

Because undef automatically acts like zero when used as a number, it's easy to make an numeric accumulator that starts out empty:

# Add up some odd numbers
$n = 1;
while ($n < 10) {
  $sum += $n;
  $n += 2; # On to the next odd number
}
print "The total was $sum.\n";

This works properly when $sum was undef before the loop started. The first time through the loop, $n is one, so the first line inside the loop adds one to $sum. That's like adding one to a variable that already holds zero (because we're using undef as if it were a number). So now it has the value 1. After that, since it's been initialized, adding works in the traditional way.

$string .= "more text\n";

If $string is undef, this will act as if it already held the empty string, putting "more text\n" into that variable. But if it already holds a string, the new text is simply appended.

Perl programmers frequently use a new variable in this way, letting it act as either zero or the empty string as needed.

Many operators return undef when the arguments are out of range or don't make sense. If you don't do anything special, you'll get a zero or a null string without major consequences. In practice, this is hardly a problem. In fact, most programmers will rely upon this behavior. But you should know that when warnings are turned on, Perl will typically warn about unusual uses of the undefined value, since that may indicate a bug. For example, simply copying undef from one variable into another isn't a problem, but trying to print it would generally cause a warning.


2.10. The while Control Structure		2.12. The defined Function