Re^2: factorial through recursion

Finally, it's very weird that your factorial function takes two inputs. That's inconsistent with the mathematical function.

Yes, but I note that his function looks similar to what a helper function would look like in a language with tail recursion optimization. It should still return a value instead of printing it though.

Comment on Re^2: factorial through recursion

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Re^3: factorial through recursion by ikegami (Patriarch) on Jul 24, 2009 at 16:10 UTC
I have no problem with a helper function having more arguments. `sub fact { my ($n) = @_; local _fact = sub { my ($n, $prod) = @_; return $prod if $n == 0; return _fact($n-1, $n$prod); }; return _fact($n, 1); } fact($n);` [download] In this case, you could even fake it as follows: `sub fact { my ($n, $prod) = @_; $prod \|\|= 1; return $prod if $n == 0; return _fact($n-1, $n$prod); } fact($n);` [download] But since Perl doesn't do tail recursion optimisation, and since sub calls are relatively slow, you're better off with: `sub fact { my ($n) = @_; my $prod = 1; $prod = $n-- while $n > 0; return $prod; } fact($n);` [download]	[reply] [d/l] [select]

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Re^3: factorial through recursion
by ikegami (Patriarch) on Jul 24, 2009 at 16:10 UTC

sub fact {
   my ($n) = @_;

   local *_fact = sub {
       my ($n, $prod) = @_;
       return $prod if $n == 0;
       return _fact($n-1, $n*$prod);
   };

   return _fact($n, 1);
}

fact($n);
[download]

In this case, you could even fake it as follows:

sub fact {
   my ($n, $prod) = @_;
   $prod ||= 1;
   return $prod if $n == 0;
   return _fact($n-1, $n*$prod);
}

fact($n);
[download]

But since Perl doesn't do tail recursion optimisation, and since sub calls are relatively slow, you're better off with:

sub fact {
   my ($n) = @_;
   my $prod = 1;
   $prod *= $n-- while $n > 0;
   return $prod;
}

fact($n);
[download]

[reply]
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