Referring to an instance method using a variable

Ruby
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 sayingproc {
|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 causesequals’ to be called. (In Python one has to use
`equals()’ to
actually perform the call).

Ideas, anybody?

Thanks,

2

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

end

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

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 W.

3

On Fri, Aug 11, 2006 at 10:56:45AM +0900, [email protected] wrote:

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.

}

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!