Re: No expansions performed while declaring an associative array using a list of keys and values
On Fri, Dec 11, 2020 at 03:08:04PM +0300, Oğuz wrote: > I was trying the new features of bash 5.1 and came across this inconsistent > behavior: > > $ foo='1 2' > $ declare -A bar=($foo 3) > $ declare -p bar > declare -A bar=(["\$foo"]="3" ) > $ > $ bar+=($foo 3) > $ declare -p bar > declare -A bar=(["\$foo"]="3" ["1 2"]="3" ) > > Is there a particular reason to avoid performing expansions in `declare -A > bar=($foo 3)'? When declaring ``associative array'', I alway use ``quoted'' syntax: $ foo='1 2' $ declare -A bar='($foo 3)' $ declare -p bar declare -A bar=(["1 2"]="3" ) Not sure how to explain this, but I'm not surprised seeing this working so. $ bar+=($foo 3) $ declare -p bar declare -A bar=(["1 2"]="3" ) Seem ok for me! -- Félix Hauri -- http://www.f-hauri.ch
Re: No expansions performed while declaring an associative array using a list of keys and values
On Mon, Dec 14, 2020 at 11:00 AM felix wrote: > On Fri, Dec 11, 2020 at 03:08:04PM +0300, Oğuz wrote: > > I was trying the new features of bash 5.1 and came across this > inconsistent > > behavior: > > > > $ foo='1 2' > > $ declare -A bar=($foo 3) > > $ declare -p bar > > declare -A bar=(["\$foo"]="3" ) > > $ > > $ bar+=($foo 3) > > $ declare -p bar > > declare -A bar=(["\$foo"]="3" ["1 2"]="3" ) > > > > Is there a particular reason to avoid performing expansions in `declare > -A > > bar=($foo 3)'? > > When declaring ``associative array'', I alway use ``quoted'' syntax: > > $ foo='1 2' > $ declare -A bar='($foo 3)' > $ declare -p bar > declare -A bar=(["1 2"]="3" ) > > Not sure how to explain this, but I'm not surprised seeing this working so. > > $ bar+=($foo 3) > $ declare -p bar > declare -A bar=(["1 2"]="3" ) > > With that it got even more confusing, see: $ foo='1 2' $ $ declare -A X=($foo 3) $ declare -p X declare -A X=(["\$foo"]="3" ) $ $ X+=($foo 3) $ declare -p X declare -A X=(["1 2"]="3" ["\$foo"]="3" ) $ $ declare -A Y='($foo 3)' $ declare -p Y declare -A Y=(["1 2"]="3" ) $ $ Y+='($foo 3)' $ declare -p Y declare -A Y=([0]="(\$foo 3)" ["1 2"]="3" ) Where did the 0 come from? > Seem ok for me! > > I think it would be the best if the only difference between declaring a variable using assignment builtins and bare assignment statements were that those builtins had the ability to set variable attributes. > -- > Félix Hauri -- http://www.f-hauri.ch > >
Re: No expansions performed while declaring an associative array using a list of keys and values
On 12/14/20 4:45 AM, Oğuz wrote: $ Y+='($foo 3)' $ declare -p Y declare -A Y=([0]="(\$foo 3)" ["1 2"]="3" ) Where did the 0 come from? That's a scalar assignment. You quoted the parens so it can't be a compound assignment. Since the variable is an array, and an assignment was performed without a subscript, you get the default subscript of "0". The difference between this and quoting the rhs of an assignment when you use `declare' is that `declare' is a builtin, and so its arguments undergo a round of expansion before `declare' sees them. That's the fundamental difference between assignment statements and arguments to declaration commands that look like assignment statements. -- ``The lyf so short, the craft so long to lerne.'' - Chaucer ``Ars longa, vita brevis'' - Hippocrates Chet Ramey, UTech, CWRUc...@case.eduhttp://tiswww.cwru.edu/~chet/
Request: a command to do a return completely in normal, ie. not subshell
Can we have a command to do a return completely (as if it returns from
main function) when it is being in a third nested function call, in
order to get back to shell prompt at once, in normal, ie. not
subshell?
My main function invocation here will fully shut down terminal if exit
command is given.
c(){
local p
#...
[ "$p" = err ] && kill $TID
#...
}
b(){
#...
c
#...
}
a(){
export TID=$$
#...
b
#...
#...
}
