Print a horizontal line¶

---- dataentry snipplet ---- snipplet_tags: terminal, line LastUpdate_dt: 2010-07-31 Contributors: Jan Schampera, prince_jammys, ccsalvesen, others type: snipplet

The purpose of this small code collection is to show some code that draws a horizontal line using as less external tools as possible (it's not a big deal to do it with AWK or Perl, but with pure or nearly-pure Bash it gets more interesting).

In general, you should be able to use this code to repeat any character or character sequence.

The simple way: Just print it¶

Not a miracle, just to be complete here.

printf '%s\n' --------------------

The iterative way¶

This one simply loops 20 times, always draws a dash, finally a newline

for ((x = 0; x < 20; x++)); do
  printf %s -
done
echo

The simple printf way¶

This one uses the printf command to print an empty field with a minimum field width of 20 characters. The text is padded with spaces, since there is no text, you get 20 spaces. The spaces are then converted to - by the tr command.

printf '%20s\n' | tr ' ' -

whitout an external command, using the (non-POSIX) substitution expansion and -v option:

printf -v res %20s
printf '%s\n' "${res// /-}"

A line across the entire width of the terminal¶

This is a variant of the above that uses tput cols to find the width of the terminal and set that number as the minimum field witdh.

printf '%*s\n' "${COLUMNS:-$(tput cols)}" '' | tr ' ' -

The more advanced printf way¶

This one is a bit tricky. The format for the printf command is %.0s, which specified a field with the maximum length of zero. After this field, printf is told to print a dash. You might remember that it's the nature of printf to repeat, if the number of conversion specifications is less than the number of given arguments. With brace expansion {1..20}, 20 arguments are given (you could easily write 1 2 3 4 ... 20, of course!). Following happens: The zero-length field plus the dash is repeated 20 times. A zero length field is, naturally, invisible. What you see is the dash, repeated 20 times.

# Note: you might see that as ''%.s'', which is a (less documented) shorthand for ''%.0s''
printf '%.0s-' {1..20}; echo

If the 20 is variable, you can use eval to insert the expansion (take care that using eval is potentially dangerous if you evaluate external data):

eval printf %.0s- '{1..'"${COLUMNS:-$(tput cols)}"\}; echo

Or restrict the length to 1 and prefix the arguments with the desired character.

eval printf %.1s '-{1..'"${COLUMNS:-$(tput cols)}"\}; echo

You can also do it the crazy ormaaj way™ following basically the same principle as this string reverse example. It completely depends on Bash due to its brace expansion evaluation order and array parameter parsing details. As above, the eval only inserts the COLUMNS expansion into the expression and isn't involved in the rest, other than to put the _ value into the environment of the _[0] expansion. This works well since we\'re not creating one set of arguments and then editing or deleting them to create another as in the previous examples.

_=- command eval printf %s '"${_[0]"{0..'"${COLUMNS:-$(tput cols)}"'}"}"'; echo

The parameter expansion way¶

Preparing enough dashes in advance, we can then use a non-POSIX subscript expansion:

hr=---------------------------------------------------------------\
----------------------------------------------------------------
printf '%s\n' "${hr:0:${COLUMNS:-$(tput cols)}}"

A more flexible approach, and also using modal terminal line-drawing characters instead of hyphens:

hr() {
  local start=$'\e(0' end=$'\e(B' line='qqqqqqqqqqqqqqqq'
  local cols=${COLUMNS:-$(tput cols)}
  while ((${#line} < cols)); do line+="$line"; done
  printf '%s%s%s\n' "$start" "${line:0:cols}" "$end"
}

printf