Hi,
I think I’ve found a nice example that shows the power of Ruby
usually not making copies of objects. But while I was trying to
take it further, I just bumped into a limitation: as far as I know it’s
impossible to replace an object with any other and changing all
the references pointing to the new one at once. It is possible to do
it with #replace(other_obj) only if the type of the new object is still
an
Array, Hash or String. But what if I want to change the underlying
type?
Let me explain what I’m trying to do in a more details:
I’m dealing with an equation system in my Ruby code, and
references are very useful in my case. Suppose that we have an
equation system that always has a variable on the left hand side,
and an expression on the right hand side, and variables have maximum
one equations where they appear on the left hand side. So, it is
straightforward to represent this in a Hash, like this:
eq = {
“k”=>12,
“y”=>[“k”, :+, 1],
“a”=>“y”,
“x”=>[“y”, :*, [“a”, :+, “k”, :+, 1]]
}
This represents the equation system
k = 12
y = k + 1
a = y
x = y * (a + k + 1)
Variables are always Strings, operators are Symbols, and compound
statements
are represented by Arrays. Now suppose that I want to substitute values
in
one
equation from another. For instance let’s replace the variable “y” in
the
equation
x = y * (…) with k+1 from the second equation (y = k+1):
eq[‘x’][0] = eq[‘y’]
The equation system changes to this:
{
“k”=>12,
“y”=>[“k”, :+, 1],
“a”=>“y”,
“x”=>[[“k”, :+, 1], :*, [“a”, :+, “k”, :+, 1]]
}
The wonderful thing in this is that instead of having a copy of it the
expression of k+1
it’s now referenced by two equations. The befit of this is that if I do
further substitutions
on the y = k+1 equation, I’ll have the modifications in the x =…
equation
too. When we
replace “k” with it’s value from the first equation:
eq[‘y’][0] = eq[‘k’] # eq[‘k’] is 12
we’ll get this
{
“k”=>12,
“y”=>[12, :+, 1],
“a”=>“y”,
“x”=>[[12, :+, 1], :*, [“a”, :+, “k”, :+, 1]]
}
Can you see the number 12 appering in the last equation? 
Awesome!
It’s
exactly what I
wanted to get.
So far so good.
Now let’s do some simplifications here. Everybody knows that 12+1 is
13,
so let’s replace
eq[‘y’] with 13. Well, the problem is, that we can’t, because eq[‘y’] =
13
will break the
carefully built reference system between the equations. The new value
of
eq[‘y’] will be
13, of course, but the other references (the one from the last equation)
pointing to the
array [12, :+, 1] will still hold the old reference to the array.
There’s one ugly solution however. I can use Array#replace on eq[‘y’]
to
replace it with
another Array, for instance with the array [13]. ([13] is an array
with a
single element that is
the numbe 13) The array [13] is not exactly 13, but very similar, and I
don’t know any
better solution at the moment:
eq[‘y’].replace [13]
This will do the trick, but [13] is not very nice, it’s like the numbe
13
in totally unnecessary
brackets:
{“k”=>12, “y”=>[13], “a”=>“y”, “x”=>[[13], :*, [“a”, :+, “k”, :+, 1]]}
I don’t want to represent 13 as [13]. What I really would love to do it
to
replace an
object (in this case the array [12, :+, 1]) with any other object (in
this
case the Fixnum
13) in a way that changes all the references from the old object to the
new
one. Just
like #replace does if the new object has the same type as the old one
(String, Array or
Hash).
Array#replace does not work with non-Array arguments, but it would be
great
to force
it to work somehow. Is it somehow possible?
u.
