Ubuntu – Unexpected EOF in a bash script, why

What you're doing is tricky. The normal way is to avoid this and just pass the array values as arguments. In order to have both options, you would have to use eval:

#!/bin/bash
function populate_array () {
    if [ "$#" -gt 0 ] ; then
        # Enter array w/ elements as argument of executable 
        # Note the quotes, they are needed
        array=("$@");
        n=$#
    else
        # Invoke executable with no arg, enter array element later
        # Read a string instead of an array and use eval to make it
        # into an array. That way, you can use tricks like escaping
        # spaces. You also need the -r option to protect the backslashes
        # so that eval will see them. 
        read -r  -p "Enter array elements separated by spaces: " string
        eval array="( $(printf '%s\n' "$string") )"
        n=${#array[@]} 
    fi
    printf "%d array elements \n" "$n"
} 

populate_array "$@"

while (("$n" > 0))   # while [ "$n" -gt 0 ] ALSO WORKS
do
    printf "%s \n" "${array[$n-1]}"
    n=$n-1
done
exit 0

You still need to escape the parentheses if you pass the array values as an argument since ( ) are reserved characters for bash. With that caveat, the script above should work as you expect:

$ foo.sh lkl1239 343.4l 3,344 \(34\) "lklk  lkl" lkaa\ lkc
6 array elements 
lkaa lkc 
lklk  lkl 
(34) 
3,344 
343.4l 
lkl1239 

And

$ foo.sh
Enter array elements separated by spaces:  lkl1239 343.4l 3,344 \(34\) "lklk  lkl" lkaa\ lkc 
6 array elements 
lkaa lkc 
lklk  lkl 
(34) 
3,344 
343.4l 
lkl1239 

The main problem is here:

read $input

In bash, usually, $foo is the value of the variable foo. Here, you don't want the value, but the name of the variable, so it should be just:

read input

Similarly, in the if tests, $yes and $no should be just yes and no, since you just want the strings yes and no there.

You could use a case statement here, which (IMHO) makes it easier to do multiple cases based on input:

case $input in
[Yy]es)    # The first character can be Y or y, so both Yes and yes work
    echo "Hello!"  
    echo "Hello!" | festival --tts
    ;;
[Nn]o)     # no or No
    echo "Are you sure?"
    echo "Are you sure?" | festival --tts
    ;;
*)         # Anything else
    echo "Please answer yes or no."
    echo "Please answer yes or no." | festival --tts
    ;;
esac

You could wrap the two echo statements and the use of festival in a function to avoid repeating yourself:

textAndSpeech ()
{
    echo "$@"
    echo "$@" | festival --tts
}
case $input in
[Yy]es)    # The first character can be Y or y, so both Yes and yes work
    textAndSpeech "Hello!"  
    ;;
[Nn]o)     # no or No
    textAndSpeech "Are you sure?"
    ;;
*)         # Anything else
    textAndSpeech "Please answer yes or no."
    ;;
esac

With $input, bash replaces this with its value, which is nothing initially, so the read command run is:

read 

And read by default stores the input in the variable REPLY. So you can, if you want, eliminate the input variable altogether and use $REPLY instead of $input.

Also have a look at the select statement in Bash.