So I'll define an indirect printing function first...
_print() while [ "$#" -ne 0 ]
do printf '$%s = %d\n' \
"$1" "$(($1))"
shift
done
Next I'll setup the array and increment...
arr=( a b c d e ); i=0
for var in "${arr[@]}"
do : "$(($var=(i+=10)))"
done
So now the value of $a is 10 and $b 20 and so on. Last there remains only to print...
_print "${arr[@]}"
...which prints to stdout...
$a = 10
$b = 20
$c = 30
$d = 40
$e = 50
All of this works because of the way the shell handles $((math)) expansions - they're basically evals. In an arithmetic expansion the shell first expands any other valid shell expansion before attempting to do the math - making the actual math its last order of business.
This means that if you do:
i=a a=10; echo "$(($i))"
The printed result is 10 because the shell expands the value of $i to get a then evaluates that result as an integer reference.
The above will work in any POSIX-compliant shell.
This means I could possibly also have done...
i=10
for var in a b c d e
do arr[($var=i*${#arr[@]}+i)/i]=$var
done
...to handle array assignment, index evaluation, and $var integer assignment at once because the [index] brackets of a named shell array - in a shell which supports them - are treated identically to the $((expansion)) parens of a math expression.
Running the above code wrapped in ksh -xc command prints the following debug output to standard error:
+ arr[1]=a
+ arr[2]=b
+ arr[3]=c
+ arr[4]=d
+ arr[5]=e
From there I can just do:
echo "$((${arr[1]}))"
echo "$((a))"
...to print...
10
10
...because they evaluate to the same thing in shells which support named arrays. However, in shells which don't...
echo 'arr=(a b c d e)' | bash -x
+ arr=(a b c d e) #bash traces the successful command to stderr
echo 'arr=(a b c d e)' | sh -x
sh: 1: Syntax error: "(" unexpected #sh traces something else
So in the _print() function I just shift over the positional parameters (which represents a truly portable shell "$@"array) while there are any at all and printf...
- First a
$ dollar sign.
- Then the
%string value stored in my $1st positional parameter.
- Then an
= equals sign.
- And last the value stored in the value stored in my
$(($1)) first positional parameter.
As the function shifts its arguments away the first positional parameter is constantly being replaced with the next until the $# count of positionals equals 0 and the function returns.
Before running the function when I initialize the array and its constituent indirection variables that works like this:
for var in "${arr[@]}"
- The shell will expand
[@] to a list of arguments and [*] to a single concatenation of that list. If the expansion is not quoted it might also expand [*] out to a list as well - depending on whether or not there is a value for $IFS when it is done and how set -filename expansion is currently configured - but if it does so it likely does not do it the way you intend.
: "$(($var=(i+=10)))"
- Each value in
${arr[@]} is assigned to the value of $var in turn. $(($var=(i+=10))) is then expanded first for the value in $var like $((a=(i+=10))) and last of all the math is done - which first increments $i by 10 and next assigns the value of $i to $a.
