P.S.:
the full test (proof, as bash is the example i compare against)
script is this:
# proof {{{
a() {
b() {
echo "b=$# 1<$1> 2<$2> 3<$3> 4<$4> 5<$5> 6<$6> 7<$7> 8<$8>
9<$9> 10<${10}> 11<${11}> 12<${12}> 13<${13}> 14<${14}> 15<${15}> 16<${16}>
17<${17}> 18<${18}> 19<${19
Date:Fri, 14 Mar 2025 01:34:48 +0100
From:Steffen Nurpmeso
Message-ID: <20250314003448.DdqG5Ont@steffen%sdaoden.eu>
| I am deeply sorry, but i have one more difficulty that i fail to
| explain to myself.
As always, there are no rules for this for MUAs so you can
Hello again.
I am deeply sorry, but i have one more difficulty that i fail to
explain to myself. It *could* (small, very small case) be that
this time the incorrectness really is on the side of bash.
This script (do not care for the commented, they are great)
a() {
echo $#,1="$1"/$1,2="$2"
On Fri, Mar 14, 2025 at 01:34:48 +0100, Steffen Nurpmeso wrote:
> a() {
> echo $#,1="$1"/$1,2="$2"/$2,3="$3"/$3,4="$4"
> echo $#,'*'="$*"/$*,
> }
> set -- '' 'a' ''
> #IFS=\ ; echo "$*"$* $*; a "$*"$* $*;unset IFS
> #IFS= ; echo "$*"$* $*; a "$*"$* $*;unset IFS
> IFS=:; echo "$*
