Referring to an instance method using a variable

ruby-talk
1

As part of a larger program I am trying to convert the following Perl code to
Ruby:

    lizzy:~% cat ptest
    sub equals {
  my($a, $b) = @_;
  return $a eq $b ? 1 : 0;
    }

    my $ops = {
  '=' => sub { my ($a, $b) = @_; return $a eq $b ? 1 : 0; },
  '==' => \&equals,
    };

    print $ops->{'='}(1, 1);
    print $ops->{'=='}(1, 2);
    lizzy:~% perl -l ptest
    1
    0
    lizzy:~%

This is what I have come up with:

    lizzy:~% cat rtest
    OPS = {
      '=' => proc { |a, b| return a == b ? 1 : 0 },
      '==' => proc { |a, b| send(:equals, a, b) },
    }

    def equals(a, b)
      return a == b ? 1 : 0
    end

    puts OPS['='].call(1, 1)
    puts OPS['=='].call(1, 2)
    lizzy:~% ruby rtest
    1
    0
    lizzy:~%

But the `==' case is rather ugly. Is there a shorter way than saying `proc {

a, b| send(:equals, a, b) }'? I.e. is there a way to avoid using the proc

wrapper?

I guess one the problems is that unlike in Python, parentheses are optional
Ruby. This means that `equals' returns what I am looking for in Python but in
Ruby it causes `equals' to be called. (In Python one has to use `equals()' to
actually perform the call).

Ideas, anybody?

Thanks,

···

--
Jos Backus
jos at catnook.com

2

Is there supposed to be a semantic difference between the '=' and '=='
operator in what you are trying to accomplish or are you just trying to
implement '=' as a proc and '==' as a method?

Does this help at all?

def equals(a, b)
   return a == b ? 1 : 0
end

OPS = {
   '=' => proc { |a, b| return a == b ? 1 : 0 },
   '==' => method(:equals),
}

puts OPS['='].call(1, 1)
puts OPS['=='].call(1, 2)

Gary Wright

···

On Aug 10, 2006, at 9:33 PM, Jos Backus wrote:

This is what I have come up with:

    lizzy:~% cat rtest
    OPS = {
      '=' => proc { |a, b| return a == b ? 1 : 0 },
      '==' => proc { |a, b| send(:equals, a, b) },
    }

    def equals(a, b)
      return a == b ? 1 : 0
    end

    puts OPS['='].call(1, 1)
    puts OPS['=='].call(1, 2)
    lizzy:~% ruby rtest
    1
    0
    lizzy:~%

3

Is there supposed to be a semantic difference between the '=' and '=='
operator in what you are trying to accomplish or are you just trying to
implement '=' as a proc and '==' as a method?

The latter. It's just an example. They are operators in a templating language
we use at work.

Does this help at all?

def equals(a, b)
  return a == b ? 1 : 0
end

OPS = {
  '=' => proc { |a, b| return a == b ? 1 : 0 },
  '==' => method(:equals),
}

puts OPS['='].call(1, 1)
puts OPS['=='].call(1, 2)

It sure does. But I could have sworn I tried that, which is why I posted :-/
Guess not.

Thanks Gary!

···

On Fri, Aug 11, 2006 at 10:56:45AM +0900, gwtmp01@mac.com wrote:

--
Jos Backus
jos at catnook.com