Question on differences on quoted $* (Was: Re: Question on $@ vs $@$@)

[Steffen Nurpmeso]

Sorry but i really have to come again, i do not understand what is
going on, and *i* think shell is at fault.

This:

  a() {
    echo $#,1=$1,2=$2,"$*",$*,
  }
  echo "$*"$* $*; a "$*"$* $*
  echo __
  IFS=:;echo "$*"$* $*; a "$*"$* $*;unset IFS
  echo ==
  set -- a b c
  echo "$*"$* $*; a "$*"$* $*
  echo 0
  IFS=:; echo "$*"$* $*; a "$*"$* $*;unset IFS

gives this output:

  1,1=,2=,,,
  __

  1,1=,2=,,,
  ==
  a b ca b c a b c
  6,1=a b ca,2=b,a b ca b c a b c,a b ca b c a b c,
  0
  a:b:ca b c a b c
  6,1=a b ca,2=b,a:b:ca:b:c:a:b:c,a b ca b c a b c,

Compared to the output of my MUA this is

  @@ -8,4 +8,4 @@ a b ca b c a b c
   6,1=a b ca,2=b,a b ca b c a b c,a b ca b c a b c,
   0
   a:b:ca b c a b c
  -6,1=a b ca,2=b,a:b:ca:b:c:a:b:c,a b ca b c a b c,
  +6,1=a:b:ca,2=b,a:b:ca:b:c:a:b:c,a b ca b c a b c,

Why does bash, when preparing the first argument to the function
a(), ie, when word-splitting "$*", not generate "a:b:c" as it very
well does when it creates the output for echo(1)?
Where is the difference in between the '"$*"$* $*' of the one
argument, and the '"$*"$* $*' of the other?
In both cases $* is in between quotes and needs to be expanded
with the first byte (character) of $IFS as a separator, this is
what i do not understand.  Is this a bug in bash?  What rule am
i missing if not?

Thank you very much.


--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)