Advanced Bash−Scripting Guide

Advanced Bash−Scripting Guide
exit 0
# Script will not exit here.
9.5. Indirect References to Variables
Assume that the value of a variable is the name of a second variable. Is it somehow possible to retrieve the
value of this second variable from the first one? For example, if a=letter_of_alphabet and
letter_of_alphabet=z, can a reference to a return z? This can indeed be done, and it is called an
indirect reference. It uses the unusual eval var1=\$$var2 notation.
Example 9−21. Indirect References
#!/bin/bash
# Indirect variable referencing.
a=letter_of_alphabet
letter_of_alphabet=z
echo
# Direct reference.
echo "a = $a"
# Indirect reference.
eval a=\$$a
echo "Now a = $a"
echo
# Now, let's try changing the second order reference.
t=table_cell_3
table_cell_3=24
echo "\"table_cell_3\" = $table_cell_3"
echo −n "dereferenced \"t\" = "; eval echo \$$t
# In this simple case,
# eval t=\$$t; echo "\"t\" = $t"
# also works (why?).
echo
t=table_cell_3
NEW_VAL=387
table_cell_3=$NEW_VAL
echo "Changing value of \"table_cell_3\" to $NEW_VAL."
echo "\"table_cell_3\" now $table_cell_3"
echo −n "dereferenced \"t\" now "; eval echo \$$t
# "eval" takes the two arguments "echo" and "\$$t" (set equal to $table_cell_3)
echo
# (Thanks, S.C., for clearing up the above behavior.)
# Another method is the ${!t} notation, discussed in "Bash, version 2" section.
# See also example "ex78.sh".
Chapter 9. Variables Revisited 94