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Re: test '-v' - associative vs. normal array discrepancy - a bug ?


From: Eduardo A . Bustamante López
Subject: Re: test '-v' - associative vs. normal array discrepancy - a bug ?
Date: Wed, 19 Nov 2014 14:41:13 -0600
User-agent: Mutt/1.5.23 (2014-03-12)

People, are we forgetting that this is supposed to work in a function, by
passing the name of a variable?

i.e. it has to look like this:

a=
b=x

is_defined a -> yes
is_defined b -> yes
is_defined c -> no

The 'length of array' expansion doesn't work when you're given a name.


So far, the best I could think of is 'is_defined3':

address@hidden ~ % ./test2
function    a   b   c   A   B   C   D   AA  BB  CC  DD
is_defined1 1   0   1   1   0   1   0   1   0   1   1
is_defined2 1   0   1   1   0   1   0   1   0   1   1
is_defined3 0   0   1   0   0   1   0   0   0   1   0


note: 0 : yes, 1 : no



Greg, close your eyes, you will *not* like this :)



address@hidden ~ % cat ./test2
#!/bin/bash

# note for is_defined[12]: if you expand an associative array with 
"address@hidden";
# it will expand to its values.

is_defined1() {
    local an="address@hidden"
    local -a a=("${!an}")

    [[ ${!1+x} = x ]] || [[ "${#a}" -gt 0 ]]
}

is_defined2() {
    local an="address@hidden"
    local -a a=("${!an}")

    [[ -v "$1" ]] || [[ "${#a}" -gt 0 ]]
}

is_defined3() {
    { declare -p -- "$1" && ! declare -fp -- "$1"; } 2>/dev/null >&2
}

a=
b=x

A=()
B=(X)
D=('')

declare -A AA=()
declare -A BB=(['x']=x)
declare -A DD=(['x']=)

declare -A is_defined{1..3}
vars=(a b c A B C D AA BB CC DD)

for var in "address@hidden"; do 
    is_defined1 "$var"; is_defined1["$var"]=$?
    is_defined2 "$var"; is_defined2["$var"]=$?
    is_defined3 "$var"; is_defined3["$var"]=$?
done

printf '%s' function
for var in "address@hidden"; do
    printf '\t%s' "$var"
done; printf \\n

for result in is_defined{1..3}; do
    printf '%s' "$result"
    for var in "address@hidden"; do
        element="$result[\"$var\"]"
        printf '\t%s' "${!element}"
    done; printf \\n
done




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