Echo character based on variable value (linux)

Question

My aim is to echo a character, for example # , based on a value such as num=6 and it must print # 6 times on the screen.

Not sure how to get this.

Answer 1

You could do something like

 printf '#%.0s' {1..6}

or, in the more general case,

 printf '#%.0s' $(seq 1 $num)

Answer 2

printf "%*s" "$num" " " | tr " " "#"

要么

yes '#' | head -"$num" | tr -d "\n"