A like your Fishing Pole analogy--a lot--but both times you brought it up, I couldn't help myself wondering why I (you) would want to use Perl to write FP?

I don't want to make Perl the next great FP language. All I want is to reduce the cost of FP in Perl so that I have the option to use FP when it's the best way to solve a problem.

Cheers,
Tom

I've hoped you would respond with some examples of how you would use this in your own code. Could you show us examples with and without it? Or at least highlight what would be different?

I'll provide some examples, but in order to appreciate them you must understand the library of tiny FP functions that form the core vocabulary of FP programming. I have enclosed a small subset of that vocabulary below, mostly drawn directly from the Haskell Prelude. I have tried to select examples that can be shown using only this subset.

Keep in mind that the following code is probably too small, too simple, and too limited to be useful for drawing general conclusions. The best you can hope for is to get a taste. To draw inferences about the whole cuisine from this tiny sampling would be a mistake.

First some library functions:

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Now let us write code using the above vocabulary. First, the preliminaries:

    use ToolBox;
    use Data::Dumper;

    # a convenience function for examining our output
  
    sub say(@) {
        local $Data::Dumper::Terse = 1;
        local $Data::Dumper::Indent = 0;
        my @d = map Dumper($_), @_;
        print "@d$/";
    }
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Let us start with some simple functions to show the general idea:

    # some simple binary operators

    sub plus  { $_[0] + $_[1] }
    sub times { $_[0] * $_[1] }

    # build n-ary operators from them

    *sum      = foldl_c( \&plus,  0 );
    *product  = foldl_c( \&times, 1 );

    # test them out

    say sum(1..10);       # 55
    say product(1..10);   # 3628800
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Now let us build further to create a function to compute vector dot products:

    *dot_product = compose( \&sum, zip_with_c( \&product ));
    say dot_product( [1,1,1], [1,2,3] ); # 6
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So far, we have computed only with numbers. Let us now turn to data. One common programming task is to compute the combinations that can be generated by taking one element from a set of sets. (I selected this problem because various implementations can be found on Perl Monks for comparison.) Here's my implementation:

    *combos = foldr_c( \&outer_prod, [[]] );

    sub outer_prod {
        my ($xs, $ys) = @_;
        [ map do { my $x=$_; map [$x, @$_], @$ys }, @$xs ];
    }


    say combos( ['a','b'], [1,2,3] );
    # [['a',1],['a',2],['a',3],['b',1],['b',2],['b',3]]
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Building further, let's compute power sets. One common method of computing the power set of a set S is the following:
1. Replace each element e of S with the set {{e},{}}:
   [1,2] ==> [ [[1],[]], [[2],[]] ]
2. Compute the combinations that can be formed from the sets:
         ==> [ [[1],[2]], [[1],[]], [[],[2]], [[],[]]] ]
3. Compute the union of the sets within each combination:
         ==> [ [1,2], [1], [2], [] ]
This method translates directly into the following code:

    *powerset = pipeline(
        map_c { [ [$_], [] ] },              # step 1
        \&combos,                            # step 2
        map_c { map [ map @$_, @$_ ], @$_ }  # step 3
    );

    say powerset( 1, 2 );
    # [1,2] [1] [2] []

    say powerset(qw( a b c ));
    # ['a','b','c'] ['a','b'] ['a','c'] ['a']
    # ['b','c'] ['b'] ['c'] []
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Each step in the original method maps directly to a stage in the composition pipeline, the output of each stage becoming the input of the next.

This ends my example.

You are welcome to code your own versions of these functions for comparison. Because these functions are so simple, the comparisons might not yield much insight. A more useful exercise might be for you to write FP and non-FP code to solve more complex problems and then compare the coding experiences instead of the code itself.

Cheers,
Tom

Tom Moertel : Blog / Talks / CPAN / LectroTest / PXSL / Coffee / Movie Rating Decoder