How to Repeat a Dash (Hyphen) in Shell

How to repeat a dash (hyphen) in shell

This throws an error:

$ printf '-%.0s' {1..100}; echo ""
bash: printf: -%: invalid option
printf: usage: printf [-v var] format [arguments]

This works fine under bash:

$ printf -- '-%.0s' {1..100}; echo ""
----------------------------------------------------------------------------------------------------

For other shells, try:

printf -- '-%.0s' $(seq 100); echo ""

The problem was the printf expects that - starts an option. As is common among Unix/POSIX utilities in this type of situation, -- signals to printf to expect no more options.

How can I repeat a character in Bash?

You can use:

printf '=%.0s' {1..100}

How this works:

Bash expands {1..100} so the command becomes:

printf '=%.0s' 1 2 3 4 ... 100

I've set printf's format to =%.0s which means that it will always print a single = no matter what argument it is given. Therefore it prints 100 =s.

How can I repeat a character in Bash?

You can use:

printf '=%.0s' {1..100}

How this works:

Bash expands {1..100} so the command becomes:

printf '=%.0s' 1 2 3 4 ... 100

I've set printf's format to =%.0s which means that it will always print a single = no matter what argument it is given. Therefore it prints 100 =s.

How to include dashes in the format string of bash printf?

Always specify end of command line flags when using strings involving --. In general the shell commands need to know where their positional argument start at. So by forcing -- after printf we let it know that the subsequent arguments are to be interpreted as its arguments. At this point, afterwards using -- will be treated literally instead of being considered as a command line switch.

so define your printf as

printf -- "--file=%s " "hello printf" "in" "bash script"

Also if you are planning to specify multiple printf argument strings,do not include them in same format specifier. You might need this

printf -- "--file=%s %s %s" "hello printf" "in" "bash script"

See more on The printf command in bash

Print a character repeatedly in bash

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
%%%%%%%%%%

Multiplying strings in bash script

You can use bash command substitution to be more portable across systems than to use a variant specific command.

$ myString=$(printf "%10s");echo ${myString// /m}           # echoes 'm' 10 times
mmmmmmmmmm

$ myString=$(printf "%10s");echo ${myString// /rep}         # echoes 'rep' 10 times
reprepreprepreprepreprepreprep

Wrapping it up in a more usable shell-function

repeatChar() {
    local input="$1"
    local count="$2"
    printf -v myString '%*s' "$count"
    printf '%s\n' "${myString// /$input}"
}

$ repeatChar str 10
strstrstrstrstrstrstrstrstrstr

---

Related Topics