bash - Looping over arrays, printing both index and value

Question

I want to do something like this:

foo=( )
foo[0]="bar"
foo[35]="baz"
for((i=0;i<${#foo[@]};i++))
do
    echo "$i: ${foo[$i]}"
done
# Output:
# 0: bar
# 1: 

Then i tried to loop through it using for in:

foo=( )
foo[0]="bar"
foo[35]="baz"
for i in ${foo[@]}
do
    echo "?: $i"
done
# Output:
# ?: bar
# ?: naz

but here I don't know the index value.

I know you could something like

foo=( )
foo[0]="bar"
foo[35]="baz"
declare -p foo
# Output:
# declare -a foo='([0]="bar" [35]="baz")'

but, can't you do it in another way?

Answer 1

You would find the array keys with "${!foo[@]}" (reference), so:

for i in "${!foo[@]}"; do 
  printf "%s%s
" "$i" "${foo[$i]}"
done

Which means that indices will be in $i while the elements themselves have to be accessed via ${foo[$i]}