Reputation: 96807
I'm interested in how one would go in getting this to work :
me = "this is a string"
class << me
alias :old<< :<<
def <<(text)
old<<(text)
puts "appended #{text}"
end
end
I'd like that when something gets appended to the me
variable, the object will use the redefined method.
If I try to run this, I get syntax error, unexpected ':', expecting kEND
at :<<
.
Upvotes: 1
Views: 790
Reputation: 868
While I agree with thenduks and ephemient, You can alias the operator that way then use send to call it, you can also still use class inheritance. e.g.:
me = "is a string"
class << me
def <<(text)
super
puts "appended #{text}"
end
end
me << " bob"
puts me #=> is a string appended bob
Upvotes: 0
Reputation: 369458
As others have already explained, the problem is simply that old<<
is not a legal Ruby identifier. You can, with tricks, create a method with that name, but you can't call it in the normal ways, and it certainly won't be recognized as an operator.
However, all the answers so far, while they have certainly answered your question, have completely ignored the underlying problem: that method shouldn't even have a name in the first place! And if it doesn't have a name, then the problem of the name being illegal simply doesn't even arise.
#!/usr/bin/env ruby
require 'test/unit'
require 'stringio'
class TestOperatorDecorator < Test::Unit::TestCase
def setup; @old_stdout, $> = $>, (@fake_stdout = StringIO.new) end
def teardown; $> = @old_stdout end
def test_that_me_dot_append_writes_to_stdio
me = 'this is a string'
class << me
old_method = instance_method :<<
define_method :<< do |text|
old_method.bind(self).(text)
puts "appended #{text}"
end
end
me << 'Test'
assert_equal "appended Test\n", @fake_stdout.string
end
end
In this case, the method never gets named, which not only means that we don't have to invent a name for it, it also means that it doesn't pollute the namespace.
Upvotes: 2
Reputation: 30442
Only certain characters are allowed in symbol literals. You are looking for:
alias :"old<<" :"<<"
Upvotes: 7
Reputation: 122439
The problem is with :old<<
. It gets interpreted as :old <<
, i.e. a symbol :old
followed by the <<
operator, so it is a syntax error. Maybe you can try :"old<<"
?
Upvotes: 1
Reputation: 204718
:old<<
looks like ":old
<<
". Try just :old
, or if you really want, :"old<<"
(but have fun calling it through that name).
Upvotes: 3