On Sat, Aug 24, 2024 at 02:15:48AM +0200, Steffen Nurpmeso wrote:
> Sorry but i really have to come again, i do not understand what is
> going on, and *i* think shell is at fault.
No. It is working correctly.
>
> This:
>
> a() {
> echo $#,1=$1,2=$2,"$*",$*,
This $1 is unquoted, it will be split. You need to use "$1" if you don't
want it to split.
> }
> echo "$*"$* $*; a "$*"$* $*
> echo __
> IFS=:;echo "$*"$* $*; a "$*"$* $*;unset IFS
> echo ==
> set -- a b c
> echo "$*"$* $*; a "$*"$* $*
> echo 0
> IFS=:; echo "$*"$* $*; a "$*"$* $*;unset IFS
>
> gives this output:
>
> 1,1=,2=,,,
> __
>
> 1,1=,2=,,,
> ==
> a b ca b c a b c
> 6,1=a b ca,2=b,a b ca b c a b c,a b ca b c a b c,
> 0
> a:b:ca b c a b c
> 6,1=a b ca,2=b,a:b:ca:b:c:a:b:c,a b ca b c a b c,
>
> Compared to the output of my MUA this is
>
> @@ -8,4 +8,4 @@ a b ca b c a b c
> 6,1=a b ca,2=b,a b ca b c a b c,a b ca b c a b c,
> 0
> a:b:ca b c a b c
> -6,1=a b ca,2=b,a:b:ca:b:c:a:b:c,a b ca b c a b c,
> +6,1=a:b:ca,2=b,a:b:ca:b:c:a:b:c,a b ca b c a b c,
>
> Why does bash, when preparing the first argument to the function
> a(), ie, when word-splitting "$*", 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?
Even though you are passing "a:b:ca" as $1, you are running echo $1 to
print it while IFS is still set to ":" so $1 will be split to "a" "b"
"ca".
> 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?
Simple example.
$ IFS=:
$ a () { echo $1 ;}
$ echo 1:2
1:2
$ a 1:2
1 2
>
> 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)
>
o/
emanuele6
