How to substitute array name with variable in korn shell?

Question

I have two arrays: options and args, and I want to display the contents of arrays. For that I wish to write a general function which will take the name of array and display its contents.

eg.

set -A options val1 val2 val3
set -A args var1 var2 var3

What I am doing now to display the contents:

dispArr() {
    i=0
    while [ "$i" -lt "${#options[*]}" ] #line1
    do
        echo ${options[$i]} #line2
        ((i=i+1))
    done
}

currently I have two different functions, one for array "options" and other for array "args"

I want to substitute array names ("options" in above code) in line1 and line2 with variables, so that I can make a call like following to print the contents of any array :

dispArr options #print the contents of array "options"
dispArr args #print the contents of array "args"

I tried providing $1 in place of array names but it didn't work. I also tried different quotes but it didn't work too.

I am new to unix and shell scripting, so any advice would be appreciated. Thanks in advance.

What version of ksh? ksh88 or ksh93? Do ksh --version -- empty output, then it's ksh88 — glenn jackman
– glenn jackman, Commented Aug 12, 2015 at 12:53
@glennjackman Following is the output of ksh --version: version sh (AT&T Research) 93u 2011-02-08 — pratz
– pratz, Commented Aug 12, 2015 at 13:20

glenn jackman · Accepted Answer · 2015-08-12 12:56:56Z

2

ksh93

dispArr() { nameref arr=$1; printf "%s\n" "${arr[@]}"; }

ksh88

dispArr() { eval printf "%s\\\\n" "\${${1}[@]}"; }

answered Aug 12, 2015 at 12:56

glenn jackman

249k42 gold badges233 silver badges362 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

How to substitute array name with variable in korn shell?

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related