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"