RogerBW's Blog

Perl Weekly Challenge 34 18 November 2019

I’ve been doing the Perl Weekly Challenges Last week’s was about demonstrating specific coding techniques.

Specifically, the call was for something to use array/hash slices, and something to use a dispatch table. I took a while to come up with something that would be well-handled by slices, but eventually settled on a moving-average calculation; each number is accompanied by the average of itself, the previous number in the list, and the next.

use List::Util qw(sum);

my @data=map {rand()} (1..10);

my @ma=map {sum(@data[$_-1..$_+1])/3} (1..$#data-1);

and similarly in Perl6 (where sum is now built in):

my @data=map {rand}, (1..10);

my @ma=map {sum(@data[$_-1..$_+1])/3}, (1..@data.end-1);

For the dispatch table I built a toy RPN calculator.

my @stack=();

my %op=(
  add => \&add,
  '+' => \&add,
  sub => \&sub,
  '-' => \&sub,
  mul => \&mul,
  '*' => \&mul,
  div => \&div,
  '/' => \&div,
  neg => \&neg,
  dup => \&dup,
  pop => \&pop,
  exch => \&exch,
    );

There's some genuine point to using a dispatch table here, because we actually have multiple references to the same thing.

Then it's just a matter of looping through the input parameters and processing them. No stack underflow detector, which is a pity.

foreach my $p (@ARGV) {
  if (exists $op{$p}) {
    $op{$p}->();
  } elsif ($p =~ /^[.0-9]+$/) {
    push @stack,$p;
  } else {
    die "Unknown input $p\n";
  }
  print join(' ',@stack),"\n";
}

And then the operators.

sub add {
  push @stack,(pop @stack) + (pop @stack);
}
sub sub {
  push @stack,-((pop @stack) - (pop @stack));
}
sub mul {
  push @stack,(pop @stack) * (pop @stack);
}
sub div {
  push @stack,1/((pop @stack) / (pop @stack));
}
sub neg {
  push @stack,-(pop @stack);
}
sub dup {
  push @stack,$stack[-1];
}
sub pop {
  pop @stack;
}
sub exch {
  ($stack[-1],$stack[-2])=($stack[-2],$stack[-1]);
}

In perl6, this is the first time I've got any kind of reference to come out right…

my @stack=();

my %op=(
  add => &add,
  '+' => &add,
  sub => &sub,
  '-' => &sub,
  mul => &mul,
  '*' => &mul,
  div => &div,
  '/' => &div,
  neg => &neg,
  dup => &dup,
  pop => &pop,
  exch => &exch,
    );

And the rest is pretty much the same except for syntax differences.

for @*ARGS -> $p {
  if (%op{$p}:exists) {
    %op{$p}.();
  } elsif ($p ~~ /^<[.0..9]>+$/) {
    push @stack,$p;
  } else {
    die "Unknown input $p\n";
  }
  say join(' ',@stack);
}

sub add {
  push @stack,(@stack.pop) + (@stack.pop);
}
sub sub {
  push @stack,-((@stack.pop) - (@stack.pop));
}
sub mul {
  push @stack,(@stack.pop) * (@stack.pop);
}
sub div {
  push @stack,1/((@stack.pop) / (@stack.pop));
}
sub neg {
  push @stack,-(@stack.pop);
}
sub dup {
  push @stack,@stack[* -1];
}
sub pop {
  @stack.pop;
}
sub exch {
  (@stack[* -1],@stack[* -2])=(@stack[* -2],@stack[* -1]);
}

I generally prefer "solve a problem by whatever means you see fit" to "demonstrate this particular technique" questions.

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