Why I Use Perl: Reliability

Post by adrianhoward » Tue, May 29, 2012, 8:50 PM UTC

jbert wrote: post #38

afc wrote: post #27

Earlier quoted context omitted.

Lets see, what about factorial in constant memory? Exponentiation (a to the power of b, where they are positive integers) in logarithmic time (not using built-in exponentiation functions/syntax)? Anonymously add a constant to a number and return it (in a way that can be passed to a function like 'map' or some such)? def Factorial(x): output = 1 for i in xrange(x): output *= (i + 1) return output def Factorial2(x): re…

Thanks for that. First one (not 100% sure it's constant memory, I think perl optimises the for to avoid instantiating the (1..$n) list: #!/usr/bin/perl use Modern::Perl; use bigint; say fact(5000); sub fact { my ($n) = @_; my $output = 1; for my $i (1..$n) { $output *= $i; } return $output; } (Took the value up to 5000 to get something which ran long enough to get a measurement, on my laptop it runs (including startu…

Post by colomon » Tue, May 29, 2012, 8:56 PM UTC

judofyr wrote: post #37

afc wrote: post #27

Earlier quoted context omitted.

Lets see, what about factorial in constant memory? Exponentiation (a to the power of b, where they are positive integers) in logarithmic time (not using built-in exponentiation functions/syntax)? Anonymously add a constant to a number and return it (in a way that can be passed to a function like 'map' or some such)? def Factorial(x): output = 1 for i in xrange(x): output *= (i + 1) return output def Factorial2(x): re…

use bignum; use v5.10; =cut def Factorial(x): output = 1 for i in xrange(x): output *= (i + 1) return output =cut # Direct translation sub factorial { my $x = shift; my $output = 1; for my $i (1 .. $x) { $output *= $i; } return $output; } say factorial(200); =cut def Factorial2(x): return reduce(operator.mul, xrange(1, x + 1), 1) =cut use List::Util "reduce"; sub factorial2 { my $x = shift; reduce { $a * $b } (1 .. $…

    sub factorial($n) {
        my $output = 1;
        for 1..$n -> $i {
            $output *= $i;
        }
        $output;
    }

    sub factorial2($n) {
        [*] 1..$n;
    }

    -> $x { $x + 7 }

    * + 7

    sub plus7($x) {
        $x + 7;
    }

Post by colomon » Tue, May 29, 2012, 8:59 PM UTC

irahul wrote: post #47

afc wrote: post #27

Earlier quoted context omitted.

Lets see, what about factorial in constant memory? Exponentiation (a to the power of b, where they are positive integers) in logarithmic time (not using built-in exponentiation functions/syntax)? Anonymously add a constant to a number and return it (in a way that can be passed to a function like 'map' or some such)? def Factorial(x): output = 1 for i in xrange(x): output *= (i + 1) return output def Factorial2(x): re…

> def Factorial(x): output = 1 for i in xrange(x): output = (i + 1) return output Umm, did you mean: def factorial(x): output = 1 for i in xrange(2, x): output *= i return output Here are the perl versions. Such trivial examples are going to look the same in languages which share the same paradigms. use List::Util qw(reduce); # This is how I would write it. sub fact1 { my $num = shift; my $output = 1; $output *= $_ f…

Post by berntb » Tue, May 29, 2012, 9:00 PM UTC

zem wrote: post #44

jbert wrote: post #38

Earlier quoted context omitted.

Thanks for that. First one (not 100% sure it's constant memory, I think perl optimises the for to avoid instantiating the (1..$n) list: #!/usr/bin/perl use Modern::Perl; use bigint; say fact(5000); sub fact { my ($n) = @_; my $output = 1; for my $i (1..$n) { $output *= $i; } return $output; } (Took the value up to 5000 to get something which ran long enough to get a measurement, on my laptop it runs (including startu…

that also highlights one of my perl peeves - having to scrape function args out of @_ rather than declaring them as part of the function definition.

  use Method::Signatures::Simple;

Post by pauljonas » Tue, May 29, 2012, 9:03 PM UTC

pwaring wrote: post #40

adrianhoward wrote: post #22

Earlier quoted context omitted.

Is COBOL still used by banks? Short answer: Yes :-) Still the backbone of many banks, building societies, credit card companies, etc. See http://www.careerjet.co.uk/cobol-jobs.html for some of job adverts. This old 2009 article http://www.guardian.co.uk/technology/2009/apr/09/cobol-inter... is still pretty much true AFAIK.

Thanks for the links - looks like the COBOL jobs are not promoted as much (I couldn't find any on the four bank sites I looked at) but still there, and reasonably well paid too.