I was talking to a coworker today about the Ruby, and we were discussing how in Ruby everything is an object. He wondered if everything was call-by-reference, and my knee jerk reaction was yes. Then I ran the following irb session.... ------------------- START IRB ------------------------- def foo(x) x = 'k' + x[1..x.size] x end def bar(x) x = 'k' x end k = 'crazy' => "crazy" foo(k) => "krazy" k => "crazy" # !?!?!?!?! bar(k) => "krazy" k => "krazy" # As expected. ------------------- END IRB ------------------------- Can someone explain to me why the method foo() doesn't modify k while bar() does? I suppose it has something to do with the fact that I'm using an asignment statement in foo() but modifiying the parameter "in-place" for bar(). Still, I'm a bit confused. [author's note: Yes, I know my example isn't very "Rubyish". I just want to get a better handle on the language.] Thanks! -Nathan
on 2006-02-24 04:59
on 2006-02-24 05:08
On Fri, Feb 24, 2006 at 12:58:38PM +0900, Nathan Morse wrote: > def foo(x) > x = 'k' + x[1..x.size] > x > end > > def bar(x) > x = 'k' > x > end String#+ creates a new object. String# edits the object in place. def foo(x) x = 'k' + x[1..x.size] [x, x.object_id] end def bar(x) x = 'k' [x, x.object_id] end >> k = 'crazy' => "crazy" >> k.object_id => 2584064 >> foo k => ["krazy", 2580574] >> bar k => ["krazy", 2584064] >> foo k => ["krazy", 2577004] >> bar k => ["krazy", 2584064] marcel
on 2006-02-24 05:38
What about when I don't use String#+ ? ------------------------------------------------------------ > def foo(x) > x = "krazy" > x > end => nil > def bar(x) > x = 'k' > x > end => nil > k = 'crazy' => "crazy" > foo(k) => "krazy" > k => "crazy" > bar(k) => "krazy" > k => "krazy"
on 2006-02-24 06:09
Nathan Morse wrote: > What about when I don't use String#+ ? Look at it this way. You're not "passing by reference." You're passing references by value. :) Learn to distinguish between variables and objects. The former are merely "labels" if you will. a = b = c = "hello" # three variables, one object Then remember that assignment always "wipes out" the old reference. Assignment doesn't change the old object at all. a = b = "hi" # a and b both refer to "hi" a = "bye" # but now a refers to "bye" So assigning a variable is different from changing an object. Look at this: a = b = "hi" b = "f" p a # "fi" We're changing the *object* referred to by b (which is the same one referred to by a). We're not making b refer to a different object. Or check this out. Both result in the value "foobar" -- but there is a difference. a = "foo" b = "foo" p a.object_id # -542440758 p b.object_id # -542444578 a << "bar" b += "bar" p a.object_id # -542440758 p b.object_id # -542466128 (a new object!) Does this help any? Hal
on 2006-02-24 07:10
<< ... remember that assignment always "wipes out" the old reference. Assignment doesn't change the old object at all. >> Ah ha! I get it now. Thanks!
on 2006-02-24 11:30
To the best of my knowledge everything (apart from integers) is passed by reference. However, 99% of the time this doesn't matter. In 'pure OOP' programming, 'messages' are sent to 'methods' and they respond with new data. If you only ever use return values you won't see inconsistent behaviour and you won't break encapsulation (which, in effect, using ByRef arguments to obtain new values does). To clarify. There are a few string methods which alter the original string object. This is also the case if you index into a string to change a char. However, when you evaluate an expression, this yields a new object so that in the following: x = x + 1 The expression on the right yields a value which is assigned to a new object, x, on the left. In other words, the x on the right is a different object from the x on the left. Since methods which alter data usually do so in the process of evaluation and assignment, most of the time any object (such as x) which "goes in" is assigned to a new object (with the same name) by the time the method ends. This explains why many people believe that arguments are sent 'by value' - because, in most cases, the value of any argument that goes into a method is not changed. Rather, a new object is created during the process of evaluation and assignment. If this is confusing, just stick to the 'pure OOP' way of doing things and always use return values; never try to use the values of ingoing arguments as 'ByRef' parameters. Life is much simpler that way ;-) best wishes Huw Collingbourne ================================ Bitwise Magazine www.bitwisemag.com Dark Neon Ltd. ================================
on 2006-02-24 14:24
Hi -- On Fri, 24 Feb 2006, Huw Collingbourne wrote: > To the best of my knowledge everything (apart from integers) is passed by > reference. Have a look at Hal's post: it's more a question of passing by value, where the values happen to be references. For example: a = "hi" meth(a) you're actually passing the value of a -- which is a reference to the string that was assigned to a. If you use a literal construct as the argument: meth("hi") then it will be wrapped in a reference for the occasion. > in the following: > argument that goes into a method is not changed. Rather, a new object is > created during the process of evaluation and assignment. > > If this is confusing, just stick to the 'pure OOP' way of doing things and > always use return values; never try to use the values of ingoing arguments > as 'ByRef' parameters. Life is much simpler that way ;-) Except that it takes you right back to the original problem -- namely, not understanding why the "values" you're manipulating are actually behaving like references :-) I'd discourage splitting the matter into the simple part and the confusing part. It's all reasonably simple, and certainly internally consistent -- and it's good to know what's actually happening, partly to avoid (or at least explain) unexpected errors, and partly to have as large a technique kit as possible should this or that need arise. David -- David A. Black (firstname.lastname@example.org) Ruby Power and Light (http://www.rubypowerandlight.com) "Ruby for Rails" chapters now available from Manning Early Access Program! http://www.manning.com/books/black