Suppose I have some bash arrays:
A1=(apple trees)
A2=(building blocks)
A3=(color television)
And index J=2, how to get the array contents of A2?
3 Answers
I've already found a resolution, this can be done by:
$ Aref=A$J
$ echo ${!Aref}
building
$ Aref=A$J[1]
$ echo ${!Aref}
blocks
$ Aref=A$J[@]
$ echo "${!Aref}"
building blocks
8 Comments
musiphil
How can you assign to, say,
A$J? You need a name without expansion to assign to, but A$J=(...) doesn't work.
Felipe Alvarez
that's the letter "A" followed by the contents of the variable "J". If J == 2, then you could see it as A + ${J} == A2
Charles Duffy
@musiphil,
printf -v "A$J" '%s' "value to assign" will do the trick. See BashFAQ #6.musiphil
@CharlesDuffy: Thanks for the pointer.
printf -v works only for a single value, not for arrays, and read -a can split a string into arrays but is subject to the delimiter and the word separator, and cannot write to an associative array. declare seems to be the most versatile: declare "A$J=hello", declare -a "A$J=(hello world)", declare -A "A$J=([x]=hello [y]=world)".Charles Duffy
@musiphil, ...back on 3.2.57,
read appears to still do the trick: read -r 'arr[1]' <<<'hello'; declare -p arr behaves as-intended. |
It’s worth to note, that even an index will be substituted at time the variable is evaluated:
$ A2=(building blocks)
$ Aref=A2[index]
$ index=1
$ echo "${!Aref}"
blocks
3 Comments
anthonyrisinger
FANTASTIC. years of shell scripting + reading the manual at least a dozen times + explicit attempts at exactly this failed to surface this little gem! note, like literal indexes, any arithmetic expression is valid, including nested expansions, eg.
cycle=(0 1 2); ref='cycle[i++%${#cycle[*]}]'; echo ${!ref} ${!ref} ${!ref} ${!ref} ${!ref} ${!ref} # => 0 1 2 0 1 2
davide
I wonder if this is a feature explicitly intended to be so, or if we are exploiting side-effect of the indirect expansion, which could silently disappear from future releases.
anthonyrisinger
@davide Still works! :-) I don't think this will change. Shell code is all about phases of macro-like literal expansion into words, and the man page has this to say: ... the first character of parameter is an exclamation point ... introduces a level of indirection ... this variable is expanded ... used in the rest of the substitution ... known as indirect expansion. AFAICT, it's 100% inline with expectations.
Today (with bash 4.3 and later), the best practice is to use nameref support:
A1=(apple trees)
A2=(building blocks)
A3=(color television)
J=2
declare -n A="A$J"
printf '%q\n' "${A[@]}"
...will properly emit:
building
blocks
This is also available as nameref A="A$J" on ksh93. See BashFAQ #6 for details.
${!ind[@]}but I never thought to introduce a temp var to solve it.d=13; e=24; f=35; a=(d e f); echo ${!a[1]}which results in "24".