Perl factorial subroutine not taking command line arguments - perl

Question

I'm trying to create a simple recursive factorial function in Perl that will take a number from the command line and then return it's factorial, e.g.
>./factorial.pl 3
>6
My subroutine doesn't seem to be taking the command line arguments. However if I take the exact same code without the sub wrapper it does take the command line arguments but obviously won't work as a subroutine. Below is the code:
#!/usr/bin/perl
use strict;
use warnings;
use diagnostics;
sub fact() {
my $number = shift or return;
return 0 if $number < 0;
my $results = 1;
while ($number--) { $results *= $number--};
return $results;
}

Answer 1

shift in a sub defaults to shifting from #_ (the sub's arguments); outside a sub, it defaults to shifting from #ARGV (the command line parameters).
So either call fact(shift) or explicitly say shift(#ARGV) in fact.
And get rid of the () prototype: sub fact {...

Answer 2

Subroutine arguments are packed in #_. Pass #ARGV to the subroutine (and get rid of the empty prototype -- do not use prototypes unless you know exactly what they do):
#!/usr/bin/env perl
use warnings; use strict;
print fact(#ARGV), "\n";
sub fact {
my ($number) = #_;
# ...
}

Answer 3

Your script has no main thread when you wrap the code in the subroutine. YOu need to actually call your subroutine, example follows.
The () in the function signature are unnecessary as you are only passing in one argument as per Sinan
#!/usr/bin/perl
use strict;
use warnings;
use diagnostics;
sub fact #() are unnecessary
{
my $number = shift or return;
return 0 if $number < 0;
my $results = 1;
while ($number--) { $results *= $number--};
return $results;
}
my $number = shift;
return fact($number);

Answer 4

The function logfac is a very accurate function of the log(factorial).
In addition.
Usually, when people ask for the factorial function, they really want it to compute a binomial coefficient. I've been using this trick since I was a student (I am only saying that so my current employer doesn't claim it as their intellectual property).
This code is numerically very accurate and very fast. Note the hardcoded limit of 20 for the "list" version... you can crank it up a little bit or tune it for performance.. I didn't bother.
For the "exact" formula, The main idea is that when computing a binomial coefficients there are a lot of cancellations. e.g. 10!/(7!*3) -> 10*9*8/(3*2*1) . To make it numerically more stable, compute this last remnant as 10/3 * 9/2 * 8/1
When the counts are too big, I used an approximation that is better converging than the stirling series. I checked up to 1000 or so.
Just call the choose m in N by 'choose(N,m)'
#Approximation from Srinivasa Ramanujan (Ramanujan 1988)
# Much better than stirling.. when tested in R.
# copyright(c) Hugues Sicotte, 2012
# Permission to use without restriction, with no implied suitability for any purposes
# use at your own risk.
use constant PI => 4 * atan2(1, 1);
sub logfac {
my $logfac=0;
my $n=shift(#_);
if($n&lt20) {
my $fac=1;
for(my $i=1;$i&lt=$n;$i++) {$fac=$fac*$i;}
$logfac=log($fac);
} else {
$logfac=n*log(n)-n+log(n*(1+4*n*(1+2*n)))/6 +log(PI)/2;
}
return $logfac;
}
sub choose {
my $total=shift #_; # N
my $choose=shift #_; # m
if($choose==0) { # N!/(0!N!) == 1, even if N==0
return 1;
} elsif($total==$choose) {#N!/N!*0!
return 1;
}
if($choose&lt20 && $total&lt20) {
my $min=$choose <$total-$choose ? $choose : $total-$choose;
my $res=$total/$min;
while($min&gt1) {
$total--;
$min--;
$res = $res * $total/$min;
}
return $res;
} else {
return exp(logfac($total)-logfac($choose)-logfac($total-$choose));
}
}

Answer 5

I've got it fixed now so that it works and it handles the case where no argument is presented.
#!/usr/bin/perl
use strict;
use warnings;
use diagnostics;
sub fact {
my $number = shift or return "Need argument!";
return 0 if $number < 0;
my $results = 1;
while ($number--) { $results *= $number--};
return $results;
}
my $number= shift;
print fact($number) . "\n";