From: Brian Candler <b.candler@...>
Date: 2008-09-19T04:17:15+09:00
Subject: Re: why one array continues to grow after repeated call

Oh I forgot to add - while playing about with this in irb, the method 
"object_id" is really useful. It returns the actual pointer. (Well, it's 
not *exactly* the pointer; it's a pointer which has been cleverly 
encoded to include some information about the type of the object, to 
optimise some common operations, but internally Ruby is able to map the 
object_id to the actual memory location with some simple bit-masking)

irb(main):001:0> a = "hello"
=> "hello"
irb(main):002:0> b = a
=> "hello"
irb(main):003:0> a.object_id
=> -605500598
irb(main):004:0> b.object_id
=> -605500598
irb(main):005:0> a << " world"
=> "hello world"
irb(main):006:0> a
=> "hello world"
irb(main):007:0> b
=> "hello world"

Here, you can see that a and b really are pointers to the same object. 
You could have done `b << " world"` instead of `a << " world"` and got 
exactly the same effect. In both cases you are invoking method "<<" with 
argument " world" on the object at "memory location"(ish) -605500598.

Contrast with:

irb(main):008:0> a = "hello"
=> "hello"
irb(main):009:0> b = "hello"
=> "hello"
irb(main):010:0> a.object_id
=> -605560548
irb(main):011:0> b.object_id
=> -605574918
irb(main):012:0> a << " world"
=> "hello world"
irb(main):013:0> a
=> "hello world"
irb(main):014:0> b
=> "hello"

You can see clearly that a and b point to different strings, which 
happen to start out with the same content.

Incidentally, if a points to a string and you want b to point to a 
different string with the same content, you can do

  b = a.dup

or

  b = String.new(a)

This creates a new String object, and copies the content byte by byte 
from the old one into the new one, leaving b pointing at the new one. In 
practice this doesn't need to be done very much.

Finally, note that if you write

  a += " world"

this is short for

  a = a + " world"

Now the expression on the RHS creates a new string object, being the 
concatenation of the original string and " world", and then this pointer 
is stored in a; a no longer points to what it did before. But assigning 
to a doesn't affect any other local variable which points to the 
original string. So:

irb(main):018:0> a = "hello"
=> "hello"
irb(main):019:0> b = a
=> "hello"
irb(main):020:0> a.object_id
=> -605642908
irb(main):021:0> b.object_id
=> -605642908
irb(main):022:0> a += " world"
=> "hello world"
irb(main):023:0> a.object_id
=> -605661638
irb(main):024:0> b.object_id
=> -605642908
irb(main):025:0> a
=> "hello world"
irb(main):026:0> b
=> "hello"

You can see that a is pointing to a new object, whereas b is still 
pointing to the original one.

I hope this makes things clearer rather than muddier :-)

Regards,

Brian.
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