Monads in Perl, with operator >>= , is it possible?

monads.pl

package Maybe;

use Moo::Role;

has value => (is => 'ro', required => 1);

use overload 
  '>>='=> \&bind;

sub bind {
  my ($self, $f) = @_;
  
  $f->($self)  
}

package Just;

use Moo;
with 'Maybe';

use overload 
    '""' => \&str,
    '+' => \∑

sub str {
  my ($self ) = @_;

  "Just(". $self->value .")"   
}

sub sum {
  my ($self, $new_value) = @_;
  
  $self->value + $new_value  
}

package main;

my $a = Just->new(value => 1);

my $f = sub { 
  my ($self ) = @_;
  Just->new(value => $self + 1)
};

my $b = $a >>= $f;

print $b; # will print Just(2)

monads2.pl

# using Attribute::Overload

package Maybe;

use Attribute::Overload;
use Moo::Role;

has value => (is => 'ro', required => 1);
 
sub bind :Overload(>>=) {
  my ($self, $f) = @_;
  
  $f->($self)  
}

package Just;

use Moo;
with 'Maybe';

sub str :Overload(""){
  my ($self ) = @_;

  "Just(". $self->value .")"   
}

sub sum :Overload(+){
  my ($self, $new_value) = @_;
  
  $self->value + $new_value  
}

package main;

my $a = Just->new(value => 1);

my $f = sub { 
  my ($self ) = @_;
  Just->new(value => $self + 1)
};

my $b = $a >>= $f;

print $b; # will print Just(2)

Interesting. Have you seen https://github.com/robrwo/perl-monads or https://github.com/hiratara/p5-Data-Monad ?

yes @robrwo but my focus was on the >>= operator, to mimic the Haskell >>=