Using overload to parse arithmetic terms

esskar has asked for the wisdom of the Perl Monks concerning the following question:

hi,

i wrote a package Foo::Expression::Arithmetic, a package Foo::Number and a package Foo::Variable. I used overload to be able to do the following:

my $x1 = Foo::Number->new()->value( 10 );
my $x2 = Foo::Number->new()->value( 20 );
my $x3 = Foo::Number->new()->value( 30 );
my $x4 = Foo::Number->new()->value( 40 );

my $v1 = Foo::Variable->new()->name( 'a' )->value( 50 );

my $c1 = $v1 - $x1 + $x2 + $x3 - $x4;
my $c2 = $c1 * $v1;
[download]

$c1 and $c2 are now objects of type Foo::Expression::Arithmetic. Foo::Expression::Arithmetic has a output function, so outputing $c1 and $c2 gives me the following:

a - 10 + 20 + 30 - 40      # $c1
a - 10 + 20 + 30 - 40 * a  # $c2

Now i want to be able to do the following:

my $c3 = ( $x1 + $x2 ) * $v1;
[download]

not achieve the following output:

( 10 + 20 ) * a   # $c3

use overload 
  '(' => \&left_parenthesis,    
  ')' => \&right_parenthesis,
[download]

is not working. :-( Any ideas?

2006-02-07 Retitled by Arunbear, as per Monastery guidelines
Original title: 'overload'

Comment on Using overload to parse arithmetic terms Select or Download Code

Replies are listed 'Best First'.
Re: Using overload to parse arithmetic terms by japhy (Canon) on Feb 07, 2006 at 12:52 UTC
If you want to output a properly parenthesized mathematical expression, you'll have to do so on the fly, examining the tree that I assume you have in your Arithmetic object. For example, if you have `$x = ($a + $b) * $c;`, then you probably have a tree like: `Arithmetic::Multiply( Arithmetic::Add( Variable('a'), Variable('b') ), Variable('c') )` [download] When you're walking that tree, when you get to a Multiply or Divide node, you have to check to see if its children are Add or Subtract nodes; if they are, those child nodes need to be surrounded by parentheses when output. Jeff `japhy` Pinyan, P.L., P.M., P.O.D, X.S.: Perl, regex, and `perl` hacker How can we ever be the sold short or the cheated, we who for every service have long ago been overpaid? ~~ Meister Eckhart	[reply] [d/l] [select]
Re^2: Using overload to parse arithmetic terms by blokhead (Monsignor) on Feb 07, 2006 at 14:41 UTC
To expand (and generalize) on this a bit, let each type of operation have a precedence. Then as you walk the expression tree, you need to parenthesize operations that have a lower precedence than the expression they appear within. A good way to do this is via OO, representing your operation types as subclasses: package Op; sub new { my $pkg = shift; bless [@_], $pkg; } sub children { @{ $_[0] } } sub display { my ($self, $precedence) = @_; $precedence = 0 if not defined $precedence; my $s = join $self->op, map { $_->display($self->precedence) } $self->children; $s = "($s)" if $precedence > $self->precedence; return $s; } package Op::Addition; @ISA = qw[Op]; sub precedence { 10 } sub op { " + " } package Op::Multiplication; @ISA = qw[Op]; sub precedence { 20 } sub op { " * " } # ... package Op::Term; @ISA = qw[Op]; sub precedence { 100 } sub display { $_[0][0] } [download] Then you just call the `display` method on an Op tree. For example, `Op::Multiplication->new( Op::Term->new(5), Op::Addition->new( Op::Term->new(6), Op::Term->new(7) ) )->display; # 5 * (6 + 7) Op::Addition->new( Op::Term->new(5), Op::Multiplication->new( Op::Term->new(6), Op::Term->new(7) ) )->display; # 5 + 6 * 7` [download] And you can see how easy it would be to add new types of operations, due to the beauty of subclassing. blokhead	[reply] [d/l] [select]
Re^3: Using overload to parse arithmetic terms by japhy (Canon) on Feb 07, 2006 at 15:38 UTC
That is a beautiful approach and technique. ++ well deserved. Jeff `japhy` Pinyan, P.L., P.M., P.O.D, X.S.: Perl, regex, and `perl` hacker How can we ever be the sold short or the cheated, we who for every service have long ago been overpaid? ~~ Meister Eckhart	[reply]
Re^3: Using overload to parse arithmetic terms by esskar (Deacon) on Feb 07, 2006 at 16:07 UTC
perfect ... thanks.	[reply]
Re: Using overload to parse arithmetic terms by Tomte (Priest) on Feb 07, 2006 at 12:30 UTC
I may be completely mistaken, but IMHO parenthesis aren't things `overload` can overload - they're at least not mentioned in overload To get better qualified help you should specify the code you wrote and mention what exactly "didn't work" in which way- your post is almost in the worst form you can ask such a question: state at least what you did, what you expected and what you got. regards, tomte An intellectual is someone whose mind watches itself. -- Albert Camus	[reply] [d/l]
Re^2: Using overload to parse arithmetic terms by esskar (Deacon) on Feb 07, 2006 at 12:43 UTC
well i, did that: `use overload '+' => \&add, '-' => \&sub, '' => \&mul;` [download] so, i'm able to build a expression that uses +, - and . Now, i want to be able to detect when the user uses parenthesis as shown above. I know that overload does not support parenthesis, but i hope that somebody has a idea how i can detect those parenthesis.	[reply] [d/l]
Re^3: Using overload to parse arithmetic terms by Tomte (Priest) on Feb 07, 2006 at 12:57 UTC
Hmm, how do you store the expressions? If you store them as a tree, you already have the information necessary, because you can always generate parenthesis for an <+\|->-Node with to children on the fly - that's what I meant with showing relevant code - do you store the expression as a tree - otherwise this is a useless info on my part g? regards, tomte An intellectual is someone whose mind watches itself. -- Albert Camus	[reply] [d/l]


more useful options
	PerlMonks