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

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 2

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

Answer 3

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 4

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

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

Answer 5

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

Firstline
Lastline

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