Question regarding sub!()


#1

Hi,

Please excuse the newbie question, but I am confused by something I see
with
sub!() (this seems to be true with other “!” operations as well):

irb(main):001:0> string1 = “hello”
=> “hello”
irb(main):002:0> string2 = string1
=> “hello”
irb(main):003:0> string2.sub!(/e/,“a”)
=> “hallo”
irb(main):004:0> string2
=> “hallo”
irb(main):005:0> string1
=> “hallo”
irb(main):006:0>

Why is string1 also changed when sub!() is being performed on string2 ?
Shouldn’t sub!() only be acting on the string it is being called on?

Thanks,
Martin


#2

when writing :

string2 = string1

both your variables reference the same address , so , modifications to
any of them will cause modification to the other .

If you write :

string2 = string1.clone

then you will modify just string2 , when writing sub!


#3

On Sun, Oct 5, 2008 at 6:37 AM, Lex W. removed_email_address@domain.invalid wrote:

then you will modify just string2 , when writing sub!

Posted via http://www.ruby-forum.com/.

But, why then, the following is not true???
n1 = 100
n2 = n2

To this point the are the same, including the object_id.
But when I assign:
n2 = 200
The n1 does not change and the object_id of n2 change.
Please see the irb session below.

irb
irb(main):001:0> n1 = 100
=> 100
irb(main):002:0> n2 = n1
=> 100
irb(main):003:0> p n1
100
=> nil
irb(main):004:0> p n2
100
=> nil
irb(main):005:0> p n1.object_id
201
=> nil
irb(main):006:0> p n2_object_id
NameError: undefined local variable or method `n2_object_id’ for
main:Object
from (irb):6
from :0
irb(main):007:0> p n2.object_id
201
=> nil
irb(main):008:0> n2 = 200
=> 200
irb(main):009:0> p n1
100
=> nil
irb(main):010:0> p n2
200
=> nil
irb(main):011:0> p n1.object_id
201
=> nil
irb(main):013:0> p n2.object_id
401
=> nil
irb(main):014:0>

Thank you

Ruby S.


#4

Hi –

On Sun, 5 Oct 2008, Ruby S. wrote:

To this point the are the same, including the object_id.
But when I assign:
n2 = 200
The n1 does not change and the object_id of n2 change.

When you assign to a variable, you break any previous binding the
variable had.

a = “string”
b = a #1
b = 100 #2

At #1, the identifiers ‘a’ and ‘b’ refer to the same object. At #2,
however, ‘b’ has been bound to another object. Nothing you do to the
string in a will affect the object in b (100).

David