A interesting obervation about object_id method

[Hank Gong]

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#integer
puts "For number"
a=3D1
b=3Da
puts a.object_id
puts b.object_id
b=3D2
puts b.object_id
b=3D1
puts b.object_id

#You can see that:
# 1) for the same integer, the object_id is the same
# 2) When b change, a will not change

#string
puts "For String"
a=3D"Hello world!"
b=3Da
puts a.object_id
puts b.object_id
b=3D"Bye world!"
puts b.object_id
b=3D"Hello world!"
puts b.object_id
b=3D"Why?"
puts b.object_id
b=3D"Why?"
puts b.object_id
puts "Why?".object_id

# 1) When you assign b=3Da, the object_id of b and a is same
# 2) but when you change b's string, the b's object_id will change
# 3) if you change b's string again, the b's object_id will also change

puts "For Array"
a=3D[1,2,3,4]
b=3Da
puts a.object_id
puts b.object_id
b[2]=3D100
puts b.object_id
puts b.join(",")
puts a.object_id
puts a.join(",")
# 1) when b=3Da, the b and a will be the same object_id
# 2) when you change b's elements, b's object_id will be same
# 3) when you change b's elements, a will also be changed

I think it's very interesting experiments. Hope someone can give us a clear
explanation.

 

[Chintan Trivedi]

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puts "For String"
a=3D"Hello world!"
b=3Da
puts a.object_id
puts b.object_id
b=3D"Bye world!"
puts b.object_id
=20
The object id of b is changing bcz u are redifining b using "=3D" oper=
ator. Instead if u use b.replace("Bye word!") thn the object id will not=
change.=20

puts "For number"
a=3D1
b=3Da
puts a.object_id
puts b.object_id
b=3D2
puts b.object_id
b=3D1
puts b.object_id

#You can see that:
# 1) for the same integer, the object_id is the same
# 2) When b change, a will not change

#string
puts "For String"
a=3D"Hello world!"
b=3Da
puts a.object_id
puts b.object_id
b=3D"Bye world!"
puts b.object_id
b=3D"Hello world!"
puts b.object_id
b=3D"Why?"
puts b.object_id
b=3D"Why?"
puts b.object_id
puts "Why?".object_id

# 1) When you assign b=3Da, the object_id of b and a is same
# 2) but when you change b's string, the b's object_id will change
# 3) if you change b's string again, the b's object_id will also change

puts "For Array"
a=3D[1,2,3,4]
b=3Da
puts a.object_id
puts b.object_id
b[2]=3D100
puts b.object_id
puts b.join(",")
puts a.object_id
puts a.join(",")
# 1) when b=3Da, the b and a will be the same object_id
# 2) when you change b's elements, b's object_id will be same
# 3) when you change b's elements, a will also be changed

I think it's very interesting experiments. Hope someone can give us a cle=
ar
explanation.

--
Huazhi Gong (Hank)



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[Alexander Jakopin]

On Wed, 2005-12-21 at 20:40 +0900, Hank Gong wrote:
[SNIP]

I think it's very interesting experiments. Hope someone can give us a clear
explanation.

A string is mutable, an integer is inmutable, that's why IMHO.
Try the same with symbolshello instead of "hello"). These are
inmutable.

 

[Dan Diebolt]

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Anyone else notice that the object_id of an integer (at least for Fixnum)=
is twice the integer plus 1?
=20
irb(main):030:0> 1.object_id
=3D> 3
irb(main):031:0> 2.object_id
=3D> 5
irb(main):032:0> 6534.object_id
=3D> 13069
irb(main):033:0> 2*6534+1
=3D> 13069
=20
=20

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[Devin Mullins]

#integer


#You can see that:
# 1) for the same integer, the object_id is the same

That's because Fixnums are "immediate." So are Symbols. It's a
performance thing. It also means, for example, you can't define
singleton methods on a Fixnum. Notice, actually, that you can decode the
integer value *from* the object-id, and vice versa.

# 2) When b change, a will not change

In Ruby, variables are *references* (a.k.a. pointers). You're just
telling b to reference a different object.

#string

# 1) When you assign b=a, the object_id of b and a is same
# 2) but when you change b's string, the b's object_id will change
# 3) if you change b's string again, the b's object_id will also change

Strings, on the other hand, are not immediate. "Hello World".object_id
!= "Hello World".object_id -- each literal reference creates a new
String object. The #replace method Chintan mentioned, however, gets
around this, by wrapping the existing String object around a new string
-- and since a and b were originally pointing to the same object, a
would change with it.

puts "For Array"
a=[1,2,3,4]
b=a
puts a.object_id
puts b.object_id
b[2]=100
puts b.object_id
puts b.join(",")
puts a.object_id
puts a.join(",")


# 1) when b=a, the b and a will be the same object_id
# 2) when you change b's elements, b's object_id will be same
# 3) when you change b's elements, a will also be changed

a and b are merely references to that one Array object, stored out in
voodoo land. The Array, itself, contains four references to other
objects. In this case, they were all Fixnums, and hence immediate (and
immutable), but try this out for size:

str = "my dog"
a=[1,2,3,4,str]
b=[:alpha, mega, str]
str['dog'] = 'cat'
p a #=> [1, 2, 3, 4, "my cat"]
p b #=> [:alpha, mega, "my cat"]

Both Arrays, while unique, reference the same String object, and so both
change at once. If, however, you did:

str = "my dog"
a=[1,2,3,4,str]
b=[:alpha, mega, str]
str = 'my cat'
p a #=> [1, 2, 3, 4, "my dog"]
p b #=> [:alpha, mega, "my dog"]

All you did was create a new String, "my cat", and point str to it. The
Arrays still point to the original "my dog" String.

I think it's very interesting experiments. Hope someone can give us a clear
explanation.

I hope that helped.

Devin

 

[Eric Hodel]

Anyone else notice that the object_id of an integer (at least for
Fixnum) is twice the integer plus 1?

irb(main):030:0> 1.object_id
=> 3
irb(main):031:0> 2.object_id
=> 5
irb(main):032:0> 6534.object_id
=> 13069
irb(main):033:0> 2*6534+1
=> 13069

Using pointers to odd memory addresses a common way of representing
integers.

non-Fixnums always have even ids since they are pointers to valid
addresses.

nil, true and false have special values.

 

[Eero Saynatkari]

That's because Fixnums are "immediate." So are Symbols. It's a
performance thing. It also means, for example, you can't define
singleton methods on a Fixnum. Notice, actually, that you can decode the
integer value *from* the object-id, and vice versa.

Even without going to implementation details, the code
is logical: each number is a constant instance of the
Fixnum (or other) class:

# Pseudo
1 = Fixnum.new("Number one")

Then, logically, all variables referencing the number
1 are pointing to the same object.

<snip comprehensive explanation />

Devin

E