Re: echo -n?

[Bob Proulx]

Alfred M. Szmidt wrote:
> Here is a fun one, how does one output `-n' (literal string) (or any
> other option that echo accepts) using echo?

Ironically 'echo' is one of the most troublesome commands for portable
usage.  You would be much better off using 'printf'.

  $ printf -- "-n\n"
  -n

  $ printf "%s\n" -n
  -n

> $ /bin/echo -- -n
> -- -n

POSIX requires that 'echo' not recognize "--" as an end of options
argument and requires that it be recognized as a string.  This is for
compatibility with traditional implementations.

  http://www.opengroup.org/onlinepubs/009695399/utilities/echo.html

> $ /bin/echo - "-n"
> - -n

Right.  Nothing special there.  Just strings.

> $ /bin/echo '-n'

The first argument is "-n".  POSIX says:

  If the first operand is -n, or if any of the operands contain a
  backslash ( '\' ) character, the results are implementation-defined.

GNU echo defines them with the BSD semantics of not outputing a newline.

> $ /bin/echo "-n"

Of course you know the shell handles argument quoting.  The command
never receives the single or double quotes.  So "-n" and '-n' are
identical in resulting behavior.

You could also use 'cat'.  A venerable old portable echo-like method
is to use cat with a here-document.

  $ cat <<EOF
  > -n
  > EOF
  -n

That is probably better than other shenanigans like:

  $ echo ' -n' | sed 's/ //'
  -n

Bob