Re: Bash bug?

[Ken Irving <fnkci@uaf.edu>]

On Thu, Mar 30, 2006 at 07:39:34PM -0500, T wrote:
> On Wed, 29 Mar 2006 20:18:07 -0600, Sumo Wrestler (or just ate too much)
> wrote:
> 
> >> the command printf seems to be ignoring the ending '\n' that passes to
> >> it:
> >> 
> >> [...]
> > 
> > It's an interaction between the echo command and the shell that's causing
> > the trailing \n to be removed. Printf is working just fine: $ printf
> > "hello\n" | od -t x1 -
> > 0000000 68 65 6c 6c 6f 0a
> > 0000006
> > This is the correct output. But what happens when you use echo: $ echo
> > -n `printf "hello\n\n\n\n\n"` | od -t x1 - 0000000 68 65 6c 6c 6f
> > 0000005
> 
> oh, thanks for the input. It really made me thinking, and I found that it
> is not echo's fault, because the previous \n was printed ok. here is more
> test:
> 
> $ echo -n "`printf "hello\n\n"`" | od -t x1 - 
> 0000000 68 65 6c 6c 6f
> 0000005
> 
> but:
> 
> $ echo -n "`printf "hello\n\n "`" | od -t x1 -
> 0000000 68 65 6c 6c 6f 0a 0a 20
> 0000010
> 
> So is it bash that is eating the trailing \n's?

Perhaps I'm missing something... but you're still using echo.

According to echo(1):

       -n     do not output the trailing newline

In the first example you indeed have trailing newlines, while in the
second there's a trailing space and so no trailing newline.  This looks
consistent to me.  

-- 
Ken Irving, fnkci@uaf.edu