Re: Re: Remove lowest number

[ellem]

First let's understand that I have no idea what I am doing and only a murky idea of what I want to do!

OK I'll give you the chomp I think that had something to do with an older version of this. See mostly I make Perl read logs and take data from different sources and shove it all together. But I don't do this everyday and as such I suck at it.

OK, so I thought:

my @num = (8, 9, 3, 5, 1, 2, 6, 7) ;
my @all = sort { $b <=> $a } @num ;

thus:

my $num5 = 5 ;
my $num4 = 4 ;
my $num3 = 3 ;
my $num2 = 2 ;

my @all = sort { $b <=> $a } $num4, $num2, $num5, $num3 ;

So when I was writing this code:

my @d1s = sort { $b <=> $a } $d61 ;

I thought I was sorting $d61 and I thought it might be hard to get at the lowest number inside $. Apparently I was wrong, but your code is over my head.

$sum += $_ for @rolls;

Doesn't mean a lot to me. The whole $_ thing confuses me.

UPDATE:

I really looked at this answer, suggestion and I started commenting things out to see which part of my code was really useless. Here's what is left:

#! /usr/local/bin/perl
use diagnostics ;
use warnings ;
use strict ;
use List::Util qw (sum) ;        #not in the standard distro

my $roll = 6 ;

until ($roll == 0) {            #rolls the dice 6 times  
    $roll -- ;
    
    my @d61 = int(rand(5) + 2) ;    #6 sided die with no 1
    my @d62 = int(rand(5) + 2) ;
    my @d63 = int(rand(5) + 2) ;
    my @d64 = int(rand(5) + 2) ;
    
    my @all = sort                 #split line for looks
    { $b <=> $a } @d61, @d62, @d63, @d64 ;    #sorts rolls 
                        #to prepare for
    pop @all ;                    #pop removing 
                            #the lowest roll
    my $sum = sum(@all) ;            #before adding
    
    print "\n@all - $sum\n" ; 
}

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