good code 4

[jimberger]

hello,

not a problem but....

can anyone think of a good efficient way of doing the following. i have three variales that are passed to a function and may or may not be populated depending on some other function. I want to check these variables and if they're empty then just assign "N/A" to the variables. One obvious way would be to do three if statements but can anyone think of a better way of doing this...

jim

 

[hmerrill]

I usually accomplish that by using the ternary(? operator:

sub some_subroutine {
my $first_arg = shift;
my $second_arg = shift;
my $third_arg = shift;

$first_arg = defined($first_arg) ? $first_arg : "N/A";
$second_arg = defined($second_arg) ? $second_arg : "N/A";
$third_arg = defined($third_arg) ? $third_arg : "N/A";

or you might want to use

$first_arg = defined($first_arg) && $first_arg ne "" ? $first_arg : "N/A";

you get the idea...

HTH. Hardy Merrill
Mission Critical Linux, Inc.

http://www.missioncriticallinux.com

 

[perlfan]

If your list ever grows beyond three the following example could do the trick as long as your assigned variables contain an alphanumeric. Note $var5 is undefined while $var2 is null.

$var1 = "a";
$var2 = "";
$var3 = "b";
$var4 = "c";
@mylist = ($var1,$var2,$var3,$var4,$var5);
@mylist = &NA_missing(@mylist);

sub NA_missing{
map{if(/\w/){$_} else {"N/A"}}@_;
}

 

[mikevh]

You could also use the ||= operator, which will assign a
new value only only if the old value is undef or zero, e.g.

my ($a, $b, $c) = map {$_ ||= "N/A"} @_;

 

[perlfan]

mikevh

Now that is clever. I thought my if else was a bit clunky but didn't see a way past it. Learned something new, thanks!

 

[mikevh]

I thought about it a little more, and you don't even need
"||=", just "||".

my ($a, $b, $c) = map {$_ || "N/A"} @_;

will do (almost) the same thing. The difference is that
the "||=" form always modifies the list that's input to
the map function, since "$_" is aliased to each consecutive
element. The second form doesn't necessarily, since it just
returns either the next list element or "N/A", whichever evaluates to "true" (not undef or zero) first. If you
assign to a different list on the left-hand side of the
map than you've got going in on the right, the original
list would be unchanged. But if "@_" held ($a, $b, $c) in
the first place (which I would assume it did), then it does change the original list, since you're assigning back into the same list.
back to the same list.