Re: escaping

[Chris F.A. Johnson]

On Wed, 3 Oct 2001, leo wrote:

> simple problem, but a basic problem nonetheless
>
> consider
>
> echo \z #backslash is a special character to the shell, so it is not printed
> result: z
> echo "\z" #when using double quotes, every special character is turned off
> except variable and argument substitution ($), backticks and backslash.
> why does the last statement output: \z instead of (what i expected) the same
> as the former command (a single z)???

>From "man bash":

        The backslash retains its special meaning only when followed by
        one of the following characters: $, `, ", \, or <newline>.

-- 
        Chris F.A. Johnson          bq933@torfree.net
        =================================================================
        c.f.a.johnson@home.com      http://cfaj.freeshell.org
        cfaj@freeshell.org          http://members.home.net/c.f.a.johnson