Print a character repeatedly in bash [duplicate]

Answer 1

sure, just use printf and a bit of bash string manipulation

$ s=$(printf "%-30s" "*")
$ echo "${s// /*}"
******************************

There should be a shorter way, but currently that's how i would do it. You can make this into a function which you can store in a library for future use

printf_new() {
 str=$1
 num=$2
 v=$(printf "%-${num}s" "$str")
 echo "${v// /*}"
}

Test run:

$ printf_new "*" 20
********************
$ printf_new "*" 10
**********
$ printf_new "%" 10
%%%%%%%%%%

Answer 2

The current accepted answer for this question (by ghostdog74) provides a method that executes extremely slowly for even a moderately high number of characters. The top-voted answer (by CaffeineConnoisseur) is better, but is still quite slow.

Here is what, in my tests, has executed fastest of all (even faster than the python version):

perl -E "print '*' x 1000"

In second place was the python command:

python -c "print('*' * 1000)"

If neither perl nor python are available, then this is third-best:

head -c 1000 /dev/zero | tr '\0' '*'

And in fourth place is the one using the bash printf built-in along with tr:

printf '%*s' 1000 | tr ' ' '*'

And here's one (from CaffeineConnoisseur's answer) that's in fifth place in speed for large numbers of repeated characters, but might be faster for small numbers (due to using only a bash built-in, and not spawning an external process):

printf '%.s*' {1..1000}

Answer 3

It's ugly, but you can do it like this:

$ for a in `seq 5`; do echo -n %; done
%%%%%

Of course, seq is an external program (which you probably have).