Re: factorial through recursion

In addition to the previously mentioned missing curlies,

my($num1,$flag)=shift; takes one value from @_ and assigns to the first of two. $flag never gets a value. You want
```
my $num1 = shift;
my $flag = shift;
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```
or
```
my ($num1, $flag) = @_;
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```
Then there's the issue that your program always dies. Return the value you find instead of dying. This will also allow you to remove the print from fact, which is appropriate since finding the factorial of a number has nothing to do with I/O.
Finally, it's very weird that your factorial function takes two inputs. That's inconsistent with the mathematical function.

Solution:

Comment on Re: factorial through recursion Select or Download Code

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Re^2: factorial through recursion by lidden (Curate) on Jul 24, 2009 at 09:36 UTC
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.	[reply]
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]