Printing n newlines with printf

Asked Dec 19 '21 at 22:42

Active Dec 19 '21 at 22:42

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I have a numeric value stored in a bash variable n and want to use the number to introduce n newlines when using printf after the arguments. How can I do this?

n=0
printf '%s\n' "$@"
printf '%s\n\n' ""

n=1
printf '%s\n' "$@"
printf '%s\n' ""

n=3
printf '%s\n' "$@"
printf '%s\n\n\n' ""

asked Dec 19 '21 at 22:42

Dilna

Hm.. while not great duplicate, most of the ideas `yes | head -n$n` `printf "%.0s\n" $(seq $n)` `perl -E 'say "\n" x '$n` from there will just work. – KamilCuk Dec 19 '21 at 22:55
`printf '%*s' $n '' | tr ' ' '\n'` – M. Nejat Aydin Dec 19 '21 at 23:08
Thought about using a loop but something is wrong in it `for (( n=1; n=nt; n++ )); do printf '%s\n' "" ; done` – Dilna Dec 19 '21 at 23:12
`for ((i=0; i – M. Nejat Aydin Dec 19 '21 at 23:14

Printing n newlines with printf

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