UNIX KSH/BASH Call by Reference using Array (at GNU bash 4.2)

Question

I am new to Unix shell scripting My code is as follows:

  i=0
for var in 'a' 'b' 'c' 'd' 'e'
do
  content[i]=var
  ((i=`expr i+1`))
done
a=10
b=50
c=40
d=90
e=100
Now I wanted to print the local variables contents using array into the function

print_array
{
echo "${content[0]}"
echo "${content[1]}"
echo "${content[2]}"
echo "${content[3]}"
echo "${content[4]}"
} 

what I wanted to do that array contents are {a,b,c,d,e} and there is a local variables named a,b,c,d,e with some assigned values I wanted to print those local variables through the use of array since the name of array element and local variable element is same so I tried to call by reference by trying multiple combinations of $ and array_names but not working

Answer 1

Try

#i=0
for var in 'a' 'b' 'c' 'd' 'e'
do
  content[${#content[*]}]=$var  # or content[i++]=$var
done
# or just content=( 'a' 'b' 'c' 'd' 'e' )
a=10
b=50
c=40
d=90
e=100
for i in "${#content[@]}"
do
  echo ${!i}
done

Answer 2

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...

1. First a $ dollar sign.
2. Then the %string value stored in my $1st positional parameter.
3. Then an = equals sign.
4. 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.

Answer 3

If I understand you correctly you want an indirection in variable's access with, e.g., bash. With this code:

content=( a b c d e )

a=10
b=50
c=40
d=90
e=100

echo "${!content[0]}"
echo "${!content[1]}"
echo "${!content[2]}"
echo "${!content[3]}"
echo "${!content[4]}"

You will get this result:

10
50
40
90
100

The key here is bash's specific variable expansion access method "${! ... }".