Re: printf "%q" and $'...'

[Greg Wooledge]

> On 25 Nov 2009, at 08:19, Antonio Macchi wrote:
> > Hi, I'm using older bash 3.2.39, so please forgiveme if in your newer bash 
> > this issue does not arise.

On Wed, Nov 25, 2009 at 08:25:00AM +0100, Maarten Billemont wrote:
> As for NUL out outputting anything in your result, the cause is C-strings.  
> Arguments are C-strings and those are delimited by NUL bytes.  Therefore, the 
> NUL byte that you're putting in it is actually marking the end of the string. 
>  So the argument ends BEFORE your NUL byte.  So it's empty.
> 
> As or \x0a, that's a newline.  And command substitution trims trailing 
> newlines.  So a string "[newline]" gets trimmed to "".

If the goal is to get content including trailing newlines into a bash
variable using printf, then Antonio can use printf -v:

imadev:~$ printf -v myvar '%q\n' $'\x0a'
imadev:~$ echo "$myvar"
$'\n'


(Note blank line in the output -- one newline from the echo command,
and one from the actual content of $myvar.)  Using printf -v instead of
x=$(printf) means you don't suffer from the trailing-newline-removal
that command substitution does.

I'm a bit puzzled by the original e-mail, though.  I don't see what the
actual goal is.  If the goal is simply "put a newline character into a
variable", then this is even simpler:

myvar=$'\n'