Bash strips “N” character when using newline-delimiter

Question

I am using Windows Linux Subsystem (Ubuntu). When I try to set a newline-delimiter, I lose my 'n'-characters. My simplified script;

#!/bin/sh
echo $HOME #gives /home/hennio
IFS=$'\n'
echo $HOME #gives /home/he  io

IFS=$'\\n\\b' didnt solve the problem. I checked my shebang with $(which sh), it is correct (although using zsh).

Searching on internet didnt give any results. Can someone please tell me whats going on? It driving me nuts..

Answer 1

To maintain compatibility, a shell invoked as /bin/sh usually tries to emulate a POSIX shell or some variant of a Bourne shell. Neither POSIX nor Bourne support $'...' . There are two possible solutions:

Method 1: Use the shebang of a shell, like bash , that supports $'...' .

Or,

Method 2: Use a POSIX method to assigning a newline to IFS :

IFS='
'

(Hat tip: Gordon Davisson )

Documentation

From man zsh :

Zsh tries to emulate sh or ksh when it is invoked as sh or ksh respectively

From man bash :

If bash is invoked with the name sh , it tries to mimic the startup behavior of historical versions of sh as closely as possible, while conforming to the POSIX standard as well.