Re: Question on $@ vs $@$@

[Steffen Nurpmeso]

Greg Wooledge wrote in
 <ZsfSX6adQ8w8tAA4@wooledge.org>:
 |On Fri, Aug 23, 2024 at 01:28:49 +0200, Steffen Nurpmeso wrote:
 |>   a() { echo $#,1=$1,2=$2,"$*",$*,; }
 |>   set -- a b c
 |>   echo 2
 |>   IFS=:; echo "$*"$*; a $* "$*";

I want to point out that bash in particular has the very dubious
description in its manual page

 If IFS has a value other than the default, then sequences of the
 whitespace characters space, tab, and newline 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  charac‐
 ter).

So IFS whitespace only if part of $IFS.

 Any  character in IFS that is not IFS whitespace, along
 with any adjacent IFS whitespace characters, delimits a field.

So this "adjacent" even if *not* part of $IFS.

 A sequence of IFS whitespace characters is also treated as
 a delimiter.

So this means that *regardless* of whatever $IFS is, the three IFS
whitespace characters are $IFS anyway *if* that is set to
a nin-empty non-default value.  I do this for things i have
documented but forgotten.  Good.

 If the  value  of IFS is null, no word splitting
 occurs.

The standard says in 2.6.5 that

  3.   A byte sequence in the input which resulted from an expansion and which
      forms an IFS white space character:
      Remove that byte sequence from the input, consider the new leading input
      byte sequence, and repeat this step.
  4.   A byte sequence in the input which resulted from an expansion and which
      forms an IFS character that is not IFS white space:
      Remove that byte sequence from the input, but note it was observed.

  At this point [after $IFS and $IFS whitespace etc], if the
  candidate is not empty, or if a sequence of bytes representing
  an IFS character that is not IFS white space was seen at step 4,
  then a field is said to have delimited, and the candidate shall
  become an output field.

So after some fresh thinking it seems this practically means what
bash documents.
I have to say i still have a lot of problems wrapping my head
against the term

  At this point [after $IFS and $IFS whitespace etc], if the
  candidate is not empty, or if a sequence of bytes representing
  an IFS character that is not IFS white space was seen at step 4,
  then a field is said to have delimited, 

It seems to me, now, that the actual point here is that IFS
whitespace can give no empty output in say a IFS=: case, whereas
the colon in $IFS *can* create empty output tokens.

Thank you.

--steffen
|
|Der Kragenbaer,                The moon bear,
|der holt sich munter           he cheerfully and one by one
|einen nach dem anderen runter  wa.ks himself off
|(By Robert Gernhardt)