Printf example in bash does not create a newline

Question

Working with printf in a bash script, adding no spaces after "\n" does not create a newline, whereas adding a space creates a newline, e. g.:

1. No space after "\n"

   NewLine=`printf "\n"`
   echo -e "Firstline${NewLine}Lastline"
   

   Result:

   FirstlineLastline
   

2. Space after "\n "

   NewLine=`printf "\n "`
   echo -e "Firstline${NewLine}Lastline"
   

   Result:

   Firstline
    Lastline
   

Question: Why doesn't 1. create the following result:

Firstline 
Lastline

I know that this specific issue could have been worked around using other techniques, but I want to focus on why 1. does not work.

Edited: When using echo instead of printf, I get the expected result, but why does printf work differently?

    NewLine=`echo "\n"`
    echo -e "Firstline${NewLine}Lastline"

Result:

    Firstline
    Lastline

Answer 1

The backtick operator removes trailing new lines. See 3.4.5. Command substitution at http://tldp.org/LDP/Bash-Beginners-Guide/html/sect_03_04.html

Note on edited question

Compare:

[alvaro@localhost ~]$ printf "\n"

[alvaro@localhost ~]$ echo "\n"
\n
[alvaro@localhost ~]$ echo -e "\n"


[alvaro@localhost ~]$

The echo command doesn't treat \n as a newline unless you tell him to do so:

NAME
       echo - display a line of text
[...]
       -e     enable interpretation of backslash escapes

POSIX 7 specifies this behaviour here:

[...] with the standard output of the command, removing sequences of one or more characters at the end of the substitution

Answer 2

Maybe people will come here with the same problem I had: echoing \n inside a code wrapped in backsticks. A little tip:

printf "astring\n"
# and 
printf "%s\n" "astring" 
# both have the same effect.
# So... I prefer the less typing one

The short answer is:

# Escape \n correctly !

# Using just: printf "$myvar\n" causes this effect inside the backsticks:
printf "banana
"

# So... you must try \\n  that will give you the desired 
printf "banana\n"

# Or even \\\\n if this string is being send to another place 
# before echoing,

buffer="${buffer}\\\\n printf \"$othervar\\\\n\""

One common problem is that if you do inside the code:

echo 'Tomato is nice'

when surrounded with backsticks will produce the error

command Tomato not found.

The workaround is to add another echo -e or printf

printed=0

function mecho(){
  #First time you need an "echo" in order bash relaxes.
  if [[ $printed == 0 ]]; then
    printf "echo -e $1\\\\n"
    printed=1
  else
    echo -e "\r\n\r$1\\\\n"
  fi
}

Now you can debug your code doing in prompt just:

(prompt)$  `mySuperFunction "arg1" "etc"`

The output will be nicely

 mydebug: a value
 otherdebug: whathever appended using myecho
 a third string

and debuging internally with

mecho "a string to be hacktyped"

Answer 3

$ printf -v NewLine "\n"
$ echo -e "Firstline${NewLine}Lastline"

Firstline
Lastline

$ echo "Firstline${NewLine}Lastline"
Firstline
Lastline

Answer 4

It looks like BASH is removing trailing newlines. e.g.

NewLine=`printf " \n\n\n"`
echo -e "Firstline${NewLine}Lastline"
Firstline Lastline

NewLine=`printf " \n\n\n "`
echo -e "Firstline${NewLine}Lastline"
Firstline


 Lastline

Answer 5

Your edited echo version is putting a literal backslash-n into the variable $NewLine which then gets interpreted by your echo -e. If you did this instead:

NewLine=$(echo -e "\n")
echo -e "Firstline${NewLine}Lastline"

your result would be the same as in case #1. To make that one work that way, you'd have to escape the backslash and put the whole thing in single quotes:

NewLine=$(printf '\\n')
echo -e "Firstline${NewLine}Lastline"

or double escape it:

NewLine=$(printf "\\\n")

Of course, you could just use printf directly or you can set your NewLine value like this:

printf "Firstline\nLastline\n"

or

NewLine=$'\n'
echo "Firstline${NewLine}Lastline"    # no need for -e

Answer 6

For people coming here wondering how to use newlines in arguments to printf, use %b instead of %s:

$> printf "a%sa" "\n"
a\na
$> printf "a%ba" "\n"
a
a

From the manual:

%b expand backslash escape sequences in the corresponding argument

Answer 7

We do not need "echo" or "printf" for creating the NewLine variable:

NewLine="
"
printf "%q\n" "${NewLine}"
echo "Firstline${NewLine}Lastline"

Answer 8

Bash delete all trailing newlines in commands substitution.

To save trailing newlines, assign printf output to the variable with printf -v VAR

instead of

NewLine=`printf "\n"`
echo -e "Firstline${NewLine}Lastline"
#FirstlineLastline

use

printf -v NewLine '\n'
echo -e "Firstline${NewLine}Lastline"
#Firstline
#Lastline

Explanation

According to bash man

3.5.4 Command Substitution

$(command)

or

`command`

Bash performs the expansion by executing command and replacing the command substitution with the standard output of the command, with any trailing newlines deleted. Embedded newlines are not deleted, but they may be removed during word splitting.

So, after adding any trailing newlines, bash will delete them.

var=$(printf '%s\n%s\n\n\n' 'foo' 'bar')
echo "$var"

output:

foo
bar

According to help printf

printf [-v var] format [arguments]

If the -v option is supplied, the output is placed into the value of the shell variable VAR rather than being sent to the standard output.

In this case, for safe copying of formatted text to the variable, use the [-v var] option:

printf -v var '%s\n%s\n\n\n' 'foo' 'bar'
echo "$var"

output:

foo
bar

Answer 9

Works ok if you add "\r"

$ nl=`printf "\n\r"` && echo "1${nl}2"
1
2