#!/usr/bin/perl w
# A nonrecursive left fold (foldl), taken from Language::Functional
sub foldl(&$$) {
my($f, $z, $xs) = @_;
map { $z = $f>($z, $_) } @{$xs};
return $z;
}
# Recursive foldl
sub foldl_rec {
my($f, $z, $xs) = @_;
my($head, @tail) = @$xs;
$head ? foldl_rec($f, $f>($z,$head), \@tail) : $z;
}
# "Fold" is the universal list traversal function. Also known as
# "reduce" (see List::Util) and "accumulate" (C++ STL). Any function
# you write that munges lists (map, grep, etc.) can be rewritten in
# terms of a fold. It essentially takes a list and replaces each "con
+s"
# constructor with a function. Stated another way, if you have a list
# @a = (1, 2, 3, 4), fold will replace the commas with another functio
+n
# of your choosing. Let's say you want the sum of the elements in @a.
# Replace the commas with a '+' sign, (1 + 2 + 3 + 4). Easy isn't it?
# You might write it as...
$s = foldl(sub{ $_[0] + $_[1] }, 0, [1..4]);
print "sum = $s\n"; # 10
# ...in addition to providing the function and the list, you supply an
# initial value to start out with. In the case of $sum above, we use
# 0. If you want the product of the elements in the list you can chan
+ge
# to...
$p = foldl(sub{ $_[0] * $_[1] }, 1, [1..4]);
print "product = $p\n"; # 24
# The "left" portion comes into play because we start at the left end
+of
# the list and work towards the right. The actual sum that is
# calculated is (((((0+1)+2)+3)+4). It only makes a difference when t
+he
# function used isn't associative. Subtraction is an example...
$l = foldl(sub{ $_[0]  $_[1] }, 0, [1..4]);
print "left fold subtraction = $l\n"; # ((((01)2)3)4) == 10
# Recursive foldr
sub foldr_rec {
my($f, $z, $xs) = @_;
my($head, @tail) = @$xs;
$head ? $f>($head,foldr_rec($f, $z, \@tail)) : $z;
}
$r = foldr_rec(sub{ $_[0]  $_[1] }, 0, [1..4]);
print "right fold subtraction = $r\n"; # (1(2(3(40)))) == 2
# The dual of "fold" is the universal list creation function, "unfold"
+.
# See more unfold in action...
#
# http://use.perl.org/~Greg%20Buchholz/journal/26747
