From: "Michael W. Ryder" <_mwryder@...>
Date: 2007-07-20T17:44:58+09:00
Subject: Re: Is there a replacement for sub?

Robert Klemme wrote:
> 2007/7/20, Michael W. Ryder <_mwryder@worldnet.att.net>:
>> Chris Shea wrote:
>> > On Jul 19, 8:59 pm, "Michael W. Ryder" <_mwry...@worldnet.att.net>
>> > wrote:
>> >> I was trying to come up with a way to remove x instances of a 
>> character
>> >> from a string and came up with a problem.  If I enter:
>> >>
>> >> a = "a b c d e f"
>> >> for i in 1..3
>> >>    a = a.sub!(' ', '')
>> >> end
>> >> puts a  ==> returns 'abcd e f' which is correct.
>> >>
>> >> But if I enter:
>> >>
>> >> a = "a b c d e f"
>> >> for i in 1..10
>> >>    a = a.sub!(' ', '')
>> >> end
>> >> puts a  ==> returns error.rb:3: private method `sub!' called for
>> >> nil:NilClass (NoMethodError, and a is now nil.
>> >>
>> >> What I am looking for is a way to remove the first n instances of a
>> >> blank from the string without wiping out the string if it does not
>> >> contain at least n blanks.  I assume there is a way to do this with
>> >> regular expressions, but have not found it yet.  This is something an
>> >> editor I liked, UCEDIT, on the CDC Cyber had in the 70's.
>> >
>> > sub! modifies the string in place, so you don't need to say a = a.sub!
>> > (' ','').  a is already changing.  And since sub! is modifying in
>> > place, it returns nil if no changes are being made, and you end up
>> > setting a to nil when that happens.
>> >
>> > a = 'a b c d e f'
>> > 10.times { a.sub!(' ','')}
>> > puts a     # 'abcdef'
>> >
>> > HTH,
>> > Chris
>> >
>>
>> I think this is where I am having problems understanding Ruby.  I have
>> to use a.sub(' ', '') in a for loop but use a.sub!(' ', '') when using a
>> times loop.  Why the difference?
> 
> There is none.  a.sub! is an alternative to a = a.sub - wherever you use 
> it.
> 

Except where you use b = a.sub!(' ', '').  Then if you loop through this 
statement 10 times b is nil, not the abcdef I expected.  This is what is 
so confusing.  At the same time I can not use b = a.sub(' ', '') as it 
always returns ab c d e f regardless of how many times I execute it, 
which is what I would expect.  So, how would I do this?
10.times { b = a.sub!(' ', '')} doesn't work, it errors out.
b = a.split(' ', 10).join works for this contrived example but I am not 
sure if it would work for something like replacing the first 12 
occurrences of \","\ with \\t\.

> robert
>