5.2. Variable Assignment

#!/bin/bash

echo

# When is a variable "naked", i.e., lacking the '$' in front?
# When it is being assigned, rather than referenced.

# Assignment
a=879
echo "The value of \"a\" is $a"

# Assignment using 'let'
let a=16+5
echo "The value of \"a\" is now $a"

echo

# In a 'for' loop (really, a type of disguised assignment)
echo -n "The values of \"a\" in the loop are "
for a in 7 8 9 11
do
  echo -n "$a "
done

echo
echo

# In a 'read' statement (also a type of assignment)
echo -n "Enter \"a\" "
read a
echo "The value of \"a\" is now $a"

echo

exit 0

#!/bin/bash

a=23              # Simple case
echo $a
b=$a
echo $b

# Now, getting a little bit fancier (command substitution).

a=`echo Hello!`   # Assigns result of 'echo' command to 'a'
echo $a
#  Note that using an exclamation mark (!) in command substitution
#+ will not work from the command line,
#+ since this triggers the Bash "history mechanism".

a=`ls -l`         # Assigns result of 'ls -l' command to 'a'
echo $a           # Unquoted, however, removes tabs and newlines.
echo
echo "$a"         # The quoted variable preserves whitespace.
                  # (See the chapter on "Quoting.")

exit 0