如何在 bash 的同一行打印多个空格然后打印 word？

Question

I need to print word with some spaces on the start of line, and I need to set spaces count with variable.

This code is not working:

spaces=20

echo "spaces: $spaces"

for ((n=0;n<$spaces;n++))
do
    printf '%s' ''
done
    
printf '%s' 'hello!'

Real output:

spaces: 20

hello!

Expected output:

spaces: 20
    
                    hello!

Answer 1

You can just use single printf to get desired number of padding of spaces:

printf '%*s\n' "$spaces" 'hello!'

               hello!

Answer 2

Change this:

    printf '%s' ''

To this:

    printf '%s' ' '

Or even this:

    printf ' '

Or just replace the whole program with:

s=20; printf %$s's\n' "hello!"