How to print whitespaces in shell scripts

Question

I am trying to implement a shell function to print a specific number of "space" characters, to provide padding for fixed-width console output.

#!/bin/bash

function printWhiteSpaces
{
  local i="122"
  local len="$1"
  local whitespace=' '
  local k=`expr $i - $len`
  while [ $k -gt 0 ]
  do
    echo "${whitespace}"
    ((k--))
  done
}

The above function takes in a single argument as a "length of something", then I take the difference of the desired field width (122) and the argument $1. And I wish to print that many whitespace characters.

Why won't the above print any whitespaces? If anyone has a solution or an improvement to the above function please share. In particular I want to know if there's a better way to write this function that prints concatenated whitespace characters.

Solution so far:

#!/bin/bash

function printWhiteSpaces
{
  local i="122"
  local len="$1"
  local k=`expr $i - $len`
  local whitespace=' '
  while [ $k -gt 0 ]
  do
    echo -n "${whitespace}"
    ((k--))
  done
}

Answer 1

If your aim is to simply print a string variable with a fixed, whitespace-padded field length, I would recommend using the printf function of bash:

printf "%122s" "$myvar"

will print the content of $myvar with as much space to the left as is necessary to fill 122 characters in total.

Try this simple test (field width taken as 20 for readability):

~> myVar="Hi, there!"; printf '%20s\n' "$myVar"
          Hi, there!
~> myVar="Bye!"; printf '%20s\n' "$myVar"
                Bye!

Your original approach didn't work because echo will implicitly print a newline at the end.

If you want to use a shell loop and only echo, you have to call it as echo -n to inhibit the newline:

~> myVar="Bye!"; l=${#myVar}; for ((i=20;i>l;i--)); do echo -n ' '; done; echo "$myVar"
                Bye!

Answer 2

This script (tested with bash and sh):

#!/bin/sh

function printWhiteSpaces() {
  local i="122"
  local len="$1"
  local k=`expr $i - $len`
  local whitespace='                                                                                                                                                                  '
  echo "${whitespace:0:$k}" test
}

printWhiteSpaces 100
printWhiteSpaces 110
printWhiteSpaces 120

Produced the following output, which I believe is what you wanted:

                       test
             test
   test

(This was done on a Mac - doubt that would make a difference though)

Answer 3

Inside the function we can tamper with the command line . So we freshly create a command line of $k number of arguments (all null). After the construction is done we echo the command line $* in double quotes will give IFS first char(= space ) separated arguments and since the arguments were all null, we will see only a sea of spaces.

function printWhitespaces
{
  local i="122"
  local len="$1"
  local whitespace=' '
  local k=`expr $i - $len`
  set -- ""
  while [ "$#" -le "$k" ]
  do
    set -- "$@" "$1"
  done
  echo "$*"
}

---

And another approach, if you already have computed the $k variable is:

printf '%*s\n' "$k" ""