Use a variable reference “inside” another variable

Question

I'm sure it is relatively simple, I just don't know how to do it.

#!/usr/bin/ksh
set `iostat`
myvar=6

I want to something like echo $($myvar) which i want interpreted as $($myvar) -> $(6) -> value

Answer 1

You can do this sort of thing with eval, built-in to many fine shells, including ksh:

#!/usr/bin/ksh
set $(iostat)
myvar=6
eval "echo \$$myvar"

The trick is to double-quote the string you feed to eval so that $myvar gets substituted with "6", and to backslash the outer dollar-sign, so that eval gets a string "$6".

I got "%user" for the output, but I tried it on a multi-processor RHEL machine.

Answer 2

In ksh93 (and recent enough versions of mksh and bash), you can write

echo ${!myvar}

Zsh requires ${(P)myvar} instead.

In older shells, including ksh88 and pdksh, your only recourse is to use eval, as explained by Bruce Ediger.