Re: why must non-standard $IFS members be treated so differently ?

[Jason Vas Dias]

Thanks Dan -
 The plot thickens - Yes, you're right, I had $IFS mistakenly set to ':' in the
  shell in which I ran 'count_args' . Without this IFS setting, I get
a count of 4:
  $ env -i PATH=/bin:/usr/bin HOME=${HOME} /bin/bash --norc
  $ count_args 1 2 3\ 4
  4
  $ IFS=: count_args 1 2 3\ 4
  3
 This to me is strange , as I've asked bash not to use ' ' as a
delimiter, when $IFS==: , but it is doing so !
 And shouldn't '3\ 4' be a single string in any case, regardless of IFS ?
 If word splitting is not doing any escaping, why not - shouldn't it
be doing so?
 Escaping works in filenames, so why not in word-splitting ?
Thanks & Regards,
Jason




On Sun, Jul 29, 2012 at 4:19 PM, Dan Douglas <ormaaj@gmail.com> wrote:
> On Sunday, July 29, 2012 03:23:29 PM Jason Vas Dias wrote:
>> echo $(count_args 1 2 3\ 4)
>
> I should also have mentioned that I couldn't reproduce this case. You should
> be getting 4 here in your example, not 3. I have the same Bash version. Are
> you sure you were echoing `${#v[@]} ' and not `${#@}', and also that you did
> not set IFS=: for count_args? If you use exactly the function you sent  with
> the default IFS then you should get 4 here.
> --
> Dan Douglas
RE:
>why must non-standard $IFS members be treated so differently ?
>Jason Vas Dias 
>3:23 PM (4 hours ago)
>
>Good day Chet, list -
> I'm concerned about the difference in output of these functions with
>the example input
> given on the '$' prefixed line below (with 4.2.29(2)-release
> (x86_64-unknown-linux-gnu)):
>
>    function count_args     {                v=($@);  echo ${#v[@]}; }
>
>    function count_colons {  IFS=':' ;    v=($@);  echo ${#v[@]}; }
>
> $ echo $(count_args 1 2 3\ 4) $(count_colons 1:2:3\:4)
> 3 4
> It appears to be impossible for an item delimited by 'X' to contain
> an escaped  'X' ('\X')  if 'X' is not
> a standard delimiter (' ', '<TAB>') .  Quoting doesn't seem to help either:
>
> $ echo $(count_args 1 2 3\ 4) $(count_colons 1:2:3':4')
> 3 4
>
>To me, this appears to be a bug.
>
>But I bet you're going to tell me it is a feature ?
>Please explain.
>
>Thanks & Regards,
>Jason
>
>BTW, documentation on $IFS does not appear to mention this issue:
>
>   Word Splitting
>       The shell scans the results of parameter expansion, command
>       substitution, and arithmetic expansion that did not occur  within
>       double quotes for word splitting.
>
>      The  shell  treats  each  character of IFS as a delimiter, and
>      splits the results of the other expansions into words on these
>     characters.  If IFS is unset, or its value is exactly
>     <space><tab><newline>, the default, then sequences of  <space>,
>     <tab>, and <newline> at the beginning and end of the results of the
>     previous expansions are ignored, and any sequence of IFS charac-
>     ters not at the beginning or end serves to delimit words.  If
>     IFS has a value other than the default, then sequences  of  the
>    whitespace  characters space and tab are ignored at the
>    beginning and end of the word, as long as the whitespace character is
>    in the value of IFS (an IFS whitespace character).  Any
>    character in IFS that is not IFS whitespace, along with any  adjacent
>    IFS whitespace characters, delimits a field.  A sequence of IFS
>   whitespace characters is also treated as a delimiter.  If the
>   value of IFS is null, no word splitting occurs.