From: dblack@...
Date: 2006-07-31T07:04:20+09:00
Subject: Re: Howto Delete 3 Leftmost Characters

Hi --

On Mon, 31 Jul 2006, Skeets wrote:

> William James wrote:
>> Skeets wrote:
>>> i'm sure this is easy, but i've gone through Pickaxe's string methods,
>>> searched the web and searched the groups, and i can't figure out how to
>>> do this in Ruby.
>>>
>>> i grep a file and it returns the following string:
>>>
>>> #ip 127.0.0.1
>>>
>>> i now want to get rid of "#ip" so i can then strip the remaining string
>>> to get rid of spaces.
>>>
>>> however, i can't find out how to delete the 3 leftmost characters - in
>>> this case "#ip".
>>>
>>> thanks for any tips to get this done - i would think it is a matter of
>>> just knowing the correct method.
>>
>> irb(main):001:0> s="#ip 127.0.0.1"
>> => "#ip 127.0.0.1"
>> irb(main):002:0> s.slice!(0,3)
>> => "#ip"
>> irb(main):003:0> s
>> => " 127.0.0.1"
>
> William, thanks.  when i follow your approach, i get your result via
> irb.  however, i get a different result when i run code in a file.  i'm
> doing something different, but i don't know what.
>
> here is the code:
>
> #!/usr/bin/env ruby
>
> if File.exist?( "ip.txt" )
>
>  f = File.open("ip.txt').grep(/#ip/)
>  (f.to_s).slice!(0,3)
> # f = f.strip
>
> end
>
> puts f  # this outputs "#ip 127.0.0.1" - i was expecting it to output
> "127.0.0.1"
>
> if i have
>
>  f = (f.to_s).slice!(0,3)
>
> instead fo
>
>  (f.to_s).slice!(0,3)
>
> then "puts f" prints "#ip"

The reason is that when you do this:

   f.to_s.slice! ...

you're operating on a string object other than f.  Ruby first does
f.to_s, which returns a new string, then does an in-place slice!
operation on that string.

In the f = ... example, you're reassigning to f, so now f is the new,
modified string.


David

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